Trigonometry

1. Trigonometric graphs

Minimum and maximum values

Given a function

\begin{aligned} f (x) = - 2 \cos x + 1 \end{aligned}

$\begin{aligned} f(x) = -2 \cos x + 1 \end{aligned}$

over the domain $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ .

What are the amplitude and range of $f(x)$ ?
Answer:
The amplitude of $f(x)$ is .

The range of $f(x)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \cos x + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(x)$ .

STEP: Determine the amplitude of $f(x)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \cos x + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \cos x + q$ is not affected by the value of $q$ .

For $f(x) = -2 \cos x + 1$ , the value of $a$ is $-\text{2}$ . As a result, the amplitude of $f(x)$ is $\text{2}$ . We ignore the sign of $-\text{2}$ as we have mentioned.

The amplitude of $f(x)$ is $\text{2}$ .

STEP: Determine the range of $f(x)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \cos x + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(x)$ .

In $f(x) = -2 \cos x + 1$ , the values and $A$ and $q$ are:
- $A = 2$
- $q = 1$
The range of $f(x)$ is:

$\begin{array}{rcl} - A + q & \leq f (x) \leq & A + q \\ - (2) + (1) & \leq f (x) \leq & (2) + (1) \\ - 1 & \leq f (x) \leq & 3 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(x) \leq & A + q \\ -(2) + (1) & \leq \; f(x)\; \leq & (2) + (1) \\ -1 & \leq \; f(x) \; \leq & 3 \end{array}$

The correct answers are:
- The amplitude is $\text{2}$ .
- The range is: $-1 \leq f(x) \leq 3$ .
Submit your answer as: and
What is the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ ?

Answer:
The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(x)$ and read off its maximum value within the given domain.

STEP: Sketch the graph of $f(x)$
[−2 points ⇒ 1 / 3 points left]

We need to find the maximum output value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ . To find this value, we need to sketch the graph of $f(x)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the maximum value from the graph.

The above figure is a sketch of the graph of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ . The maximum value has been marked with a dot.

STEP: Read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ directly from the graph. The coordinates of the maximum value are $(\text{180}\text{°}; \text{3})$ . The maximum value is the $y$ -value of the coordinates $(\text{180}\text{°}; \text{3})$ , that is, $\text{3}$ .

The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ is $\text{3}$ .

Submit your answer as:

Finding a cosine equation from a graph.

The graph below shows a trigonometric equation of the following form: $y = a \cos x + q$ graphed over the domain $x \in \left[ 0° ; 360° \right]$ . Two points are shown on the graph: Point A at $\left(270°;\text{1} \right)$ , and point B at $\left(360°;\text{4} \right)$ .

Calculate the values of $a$ (the amplitude of the graph) and $q$ (the vertical shift of the graph).

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Look at the points given: are they at the peaks (maximum values) of the cosine curve or at the troughs (minimum values)? Or are they in the 'middle' of the graph? Use these points to decide how tall the graph is: from that you can find the amplitude of the graph,

a

$a$ .

STEP: Determine the amplitude of the curve
[−2 points ⇒ 2 / 4 points left]

There are some important points on a cosine graph. These are the maximum points, the minimum points, and the points in the middle. With those points we can figure out the values of $a$ and $q$ .

The graph below shows the maximum, minimum and middle values of the equation in this question. The sinusoidal axis, which is the line in the middle, sits halfway from the top to the bottom. In other words, it is exactly halfway between the maximum and minimum! The picture also shows the amplitude of the graph. It is the distance from the middle to the maximum. Note that the amplitude is always positive because it is a distance.

In this question, we have one point at the middle and another point at the top.

Let's start by finding the amplitude of the graph, which is the value of $a$ in the equation. The $y$ -value at the middle is $\text{1}$ , while the $y$ -value at the the top is $\text{4}$ . We can find the amplitude by working out the distance from the middle of the graph to the top: $\text{4} - (\text{1}) = 3$ .

At this point, we know that the equation is $y = 3 \cos x + q$ .

STEP: Determine the vertical shift of the curve
[−2 points ⇒ 0 / 4 points left]

Now on to finding the value of $q$ . This can actually be done very quickly. Normally the sinusiodal axis sits at $y = 0$ . But the $q$ term moves that axis up or down. In this graph, the sinusoidal axis is at $y = 1$ ; the graph has been shifted up 1 space. Therefore $q = 1$ .

The equation for the graph shown is $y = 3 \cos x+ 1$ .

The correct answers are $a = 3$ and $q = \text{1}$ .

Submit your answer as: and

Calculating the period of sine graphs

Shown the following graph of the following form: $f(x) = \sin(kx)$ .

Calculate the value of $k$ .

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \sin (k x)

$f(x) = \sin(kx)$ with the period of

y = \sin (x)

$y = \sin(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \sin(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{270} \\ scale & = \frac{4}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{270}} \\ \textrm{scale} & = \frac{4}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{4}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $\frac{4}{3}$ .

NOTE:The final equation of the graph is:

f (x) = \sin (\frac{4}{3} \cdot x)

$f(x) = \sin(\frac{4}{3} \cdot x)$

Submit your answer as:

Finding a tangent equation from a graph

The graph below shows a tangent curve with an equation of the form $y = a \tan x + q$ . Two points are labelled on the curve: Point A is at $\left( 0° ; -1 \right)$ , and point B is at $\left( 315° ; -4 \right)$ .

Find $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The tangent graph normally crosses through the origin,

(0 °; 0)

$(0°;0)$ . However, the value of

q

$q$ shifts the graph up or down. Use point

A

$A$ to figure out how much the graph has moved up or down. That tells you the value of

q

$q$ .

STEP: Read off the value of

q

$q$ from point

A

$A$
[−1 point ⇒ 2 / 3 points left]

The graph of tangent normally has $x$ -intercepts at $x = 0°, 180°$ and $360°$ . The $q$ term in the equation has the effect of shifting the graph up or down. We can use the position of point $A$ to figure out how much the graph moved up or down: since point $A$ is at $\left( 0° ; -1 \right)$ , the graph moved down by $1$ space. So $c = -1$ .

At this point we know that the equation is $y = a \tan x -1$ .

STEP: Substitute in the coordinates of a point and solve for

a

$a$
[−2 points ⇒ 0 / 3 points left]

To work out the value of $a$ , we will use a method which is very common: we will substitute coordinates from a point on the graph into the equation and then solve for $a$ . It is important to understand that this method works because every point on the curve for the equation 'agrees' with the equation. If you put the coordinates into the equation, it must work! We will use the coordinates of point $B$ , $\left( 315° ; -4 \right)$ .

\begin{aligned} y & = a \tan x - 1 \\ _{coordinates of point B}^{substitute in the} ⟶ (- 4) & = a \tan (315 °) - 1 \\ - 4 & = a (- 1) - 1 \\ - 4 + 1 & = - a \\ - 3 & = - a \\ 3 & = a \end{aligned}

$\begin{aligned} y & = a \tan x -1 \\ ^{\text{substitute in the}}_{\text{coordinates of point }B} \quad \longrightarrow \quad \left( -4 \right) & = a \tan \left( \text{315}° \right) -1 \\ -4 & = a \left( -\text{1} \right) -1 \\ -4 + 1 & = - a \\ -3 & = - a \\ 3 & = a \end{aligned}$

There we have it: the value of $a$ is $3$ . Putting it all together, we have the complete equation, which is $y = 3 \tan x -1$ .

The answers are $a = 3$ and $q = -1$ .

Submit your answer as: and

Finding a sine equation from a graph

The graph here shows an equation of the following form: $y = a \sin x + q$ . Point $A$ is at $(360°;\text{1,5})$ , and point $B$ is at $(180°;\text{1,5})$ . Calculate the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The value of $a$ is linked to the height of the sine curve, while the value of $q$ shifts the curve up or down.

STEP: Use the amplitude to find the value of

a

$a$
[−1 point ⇒ 1 / 2 points left]

The most important places on the graph of a sine curve (and also the cosine curve) are the maximum and minimum values, as well as the middle of the graph. These key places on the graph are shown below. Notice that the amplitude is the distance from the centre line of the graph - called the 'sinusoidal axis' - and the peak value; the amplitude tells how tall the graph is. (It is important to remember that the amplitude of the curve is always positive, because it is a distance.)

In this question, we have one point at the middle and another point at the middle of the sine curve. We can use those points to find both $a$ and $q$ .

Although we can find either of the two answers from the information given in the question, we will figure out the answer for $a$ first: $a$ is directly related to the amplitude of the curve. The $y$ -value at the middle is $\text{1,5}$ , while the $y$ -value at the middle is $\text{1,5}$ . We can find the amplitude by working out the distance from the middle of the graph to the middle: $\text{1,5} - (\text{1,5}) = \frac{1}{2}$ .

At this point, we have worked out half of the answer: the equation is $y = \frac{1}{2} \sin x + q$ .

STEP: Find

q

$q$ from the vertical shift of the curve
[−1 point ⇒ 0 / 2 points left]

Now we will figure out the value of $q$ . Remember that the sinusoidal axis for sine and cosine usually sits at $y = 0$ ; but the $q$ term moves that axis up or down. Look at the graph to see how the curve moved: in this graph, the sinusoidal axis is at $y = \frac{3}{2}$ ; the graph has been shifted up $\frac{3}{2}$ spaces. So $q = \frac{3}{2}$ .

The complete equation for the graph shown in this question is $y = \frac{1}{2} \sin x+ \frac{3}{2}$ .

The correct answers are:

$a = \frac{1}{2}$

$q = \frac{3}{2}$

Submit your answer as: and

Working with the period of tangent graphs

The graph below shows a tangent function of the following form: $f(x) = \tan(kx)$ . The asymptotes lie at $-\text{135}°$ and $\text{135}°$ .

Calculate the value of $k$ .

INSTRUCTION: Your answer must be either an integer or a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \tan (k x)

$f(x) = \tan(kx)$ with the period of

y = \tan (x)

$y = \tan(x)$
[−1 point ⇒ 2 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \tan(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{90}{135} \\ scale & = \frac{2}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{90}{\text{135}} \\ \textrm{scale} & = \frac{2}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{2}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $\frac{2}{3}$ .

NOTE: The final equation of the graph is therefore:

f (x) = \tan (\frac{2}{3} \cdot x)

$f(x) = \tan(\frac{2}{3} \cdot x)$

Submit your answer as:

The cosine function

The table below shows input and output values for the trigonometric function $f(x) = \cos x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$1$
$60°$	$\frac{1}{2}$
$120°$	$-\frac{1}{2}$
$180°$	$-1$
$240°$	$-\frac{1}{2}$
$300°$	$?$
$360°$	$1$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \cos x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(300°)$ . So we need to evaluate the function with $x = 300°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(300°)$ is $\frac{1}{2}$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $\dfrac{1}{2}$ .

Submit your answer as:

Properties of sine functions

The diagram below shows the graph of $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the peak of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depth of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a peak: $(\text{90}\text{°}; \text{2})$ . We will substitute these coordinates into the equation of $g(x)$ and solve for $a$ .

$\begin{aligned} g (x) & = a \sin x \\ (2) & = a \sin (90 °) \\ 2 & = a (1) \\ 2 & = a \end{aligned}$ $\begin{aligned} g(x) &= a \sin x \\ (2) &= a \sin (90°) \\ 2 &= a (1) \\ 2 &= a \end{aligned}$

NOTE: If we look at the graph of $g(x)$ again, we see that the height of the peaks or depth of the troughs is $\text{2}$ . This means that the amplitude of $g(x)$ is $\text{2}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any sine function of the form $f = a \sin x$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(x)$ : $g(x) = 2 \sin x$ .

The value of $a$ is $\text{2}$ .

Submit your answer as:
The graph of $g$ is translated $\text{4}$ units downwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 2\sin x + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 2\sin x + q$ , where $q$ is the vertical shift. The value of $q$ is positive if the graph is shifted upwards and negative if the graph is shifted downwards.

The graph of $g$ has been shifted downwards by $\text{4}$ units. This means the value of $q$ is $-\text{4}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ downwards by $\text{4}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 2$
- $q = -4$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (2) + (- 4) & \leq h \leq & (2) + (- 4) \\ - 6 & \leq h \leq & - 2 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(2) + (-4) & \leq \; h \; \leq & (2) + (-4) \\ -6 & \leq \; h \; \leq & -2 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $-\text{6}$ and the maximum is $-\text{2}$ .

The correct answer is: $-6 \leq h \leq -2$ .

Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function $f(x) = \tan x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$45°$	$?$
$90°$	$\text{undefined}$
$135°$	$-1$
$180°$	$0$
$225°$	$1$
$270°$	$\text{undefined}$
$315°$	$-1$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \tan x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(45°)$ . So we need to evaluate the function with $x = 45°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(45°)$ is $1$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $1$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (θ) = - 2 \sin θ - 1 \end{aligned}

$\begin{aligned} f(\theta) = -2 \sin \theta - 1 \end{aligned}$

over the domain $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ .

What are the amplitude and range of $f(\theta)$ ?
Answer:
The amplitude of $f(\theta)$ is .

The range of $f(\theta)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \sin \theta + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(\theta)$ .

STEP: Determine the amplitude of $f(\theta)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \sin \theta + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \sin \theta + q$ is not affected by the value of $q$ .

For $f(\theta) = -2 \sin \theta - 1$ , the value of $a$ is $-\text{2}$ . As a result, the amplitude of $f(\theta)$ is $\text{2}$ . We ignore the sign of $-\text{2}$ as we have mentioned.

The amplitude of $f(\theta)$ is $\text{2}$ .

STEP: Determine the range of $f(\theta)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \sin \theta + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(\theta)$ .

In $f(\theta) = -2 \sin \theta - 1$ , the values and $A$ and $q$ are:
- $A = 2$
- $q = -1$
The range of $f(\theta)$ is:

$\begin{array}{rcl} - A + q & \leq f (θ) \leq & A + q \\ - (2) + (- 1) & \leq f (θ) \leq & (2) + (- 1) \\ - 3 & \leq f (θ) \leq & 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(\theta) \leq & A + q \\ -(2) + (-1) & \leq \; f(\theta)\; \leq & (2) + (-1) \\ -3 & \leq \; f(\theta) \; \leq & 1 \end{array}$

The correct answers are:
- The amplitude is $\text{2}$ .
- The range is: $-3 \leq f(\theta) \leq 1$ .
Submit your answer as: and
What is the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ ?

Answer:
The minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(\theta)$ and read off its minimum value within the given domain.

STEP: Sketch the graph of $f(\theta)$
[−2 points ⇒ 1 / 3 points left]

We need to find the minimum output value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ . To find this value, we need to sketch the graph of $f(\theta)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the minimum value from the graph.

The above figure is a sketch of the graph of $f(\theta)$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ . The minimum value has been marked with a dot.

STEP: Read off the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ directly from the graph. The coordinates of the minimum value are $(\text{90}\text{°}; -\text{3})$ . The minimum value is the $y$ -value of the coordinates $(\text{90}\text{°}; -\text{3})$ , that is, $-\text{3}$ .

The minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ is $-\text{3}$ .

Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function $f(x) = \sin x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$?$
$90°$	$1$
$180°$	$0$
$270°$	$-1$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \sin x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(0°)$ . So we need to evaluate the function with $x = 0°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(0°)$ is $0$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $0$ .

Submit your answer as:

Trigonometry: interpreting graphs

The graphs show the functions $f(x) = c\cos x + d$ and $g(x) = a\tan x + b$ . Points $A( \text{45}; \text{0,5})$ and $B( \text{0}; \text{3})$ are labelled, and Point $N (x;y)$ is on $g(x)$ .

Find the numerical values of $a$ and $b$ for $g(x)$ .
Answer:
1. $a =$
2. $b =$
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Begin by identifying which graph is $f(x)$ and which is $g(x)$ . What are the effects of $a$ and $b$ on the graph?

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

It is often easiest to first find the vertical shift of the graph, $b$ . Examine the graph of $g(x)$ and determine if it has gone up or down. The graph in question is shown below highlighted in blue, and the centre line is shown as well. The centre line has not moved up or down from the $x$ -axis. Therefore, the value of $b$ is $\text{0}$ .

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The value of $a$ affects the amplitude: either stretching or flattening the graph. You can find the amplitude by looking at the peaks and valleys of the graph. For $\tan x$ it is important to remember that the normal graph passes through the point $(45° ; 1)$ . If the $\tan$ graph is multiplied by a coefficient other than one, it will move that point up or down by stretching or squeezing it. Making these comparisons, you should find that $a = \frac{1}{2}.$

Submit your answer as: and
What is the maximum value of $g(x) -1$ ?

Answer: The answer is .
expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Questions about 'minimum' or 'maximum' values are about the lowest and highest points on a graph (the smallest and biggest $y$ -values). Think about the effect of the $-1$ on the graph: will it move the graph horizontally or vertically?

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

This question asks you to think about the graph of $g(x)$ when it has been changed due to the extra term of $-1$ . This term moves the entire graph down $-1$ space(s). In this case, you are moving the $\tan$ function, which goes down forever and up forever (along the asymptotes): the minimum value is $-\infty$ and the maximum value is $\infty$ . If the entire graph moves down $-1$ space(s), the graph still reaches down to $-\infty$ and up to $\infty$ . Therefore, the maximum value of $g(x) -1$ is: $\infty$ .

Submit your answer as:
Determine the range of $g(x) -1$ .

INSTRUCTION: Write your answer in interval notation.

Answer: $y \in$
interval

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The range of a graph is the total 'height' of the graph from the smallest $y$ -value up to the maximum $y$ -value.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The range of the graph is the interval from the lowest $y$ -value up to the largest $y$ -value. In the case of $g(x) -1$ , it is the graph of $g(x)$ shifted down by $\text{1}$ space(s). You can see on the graph that $g(x)$ has a minimum value of $y = -\infty$ and a maximum value of $y = \infty$ . When the graph is shifted due to the $-1$ , the range is the interval $(-\infty; \infty)$ .

Submit your answer as:
Given $\cos(\alpha) = k$ , find a simplified expression for $\cos(\alpha -90°)$ in terms of $\alpha$ and $k$ .

INSTRUCTION: Type a in the place of $\alpha$ , and round your answer to two decimal places if there are non-integer values.

Answer: $\cos(\alpha -90°) =$
expression

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
This question is about compound angles: you need to use the compound angle identity which matches the question to expand the expression and get an answer.

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Simplifying the expression in the question requires using the compound angle identity $\cos \left( A - B\right) = \cos A \cos B + \sin A \sin B$ to expand the function by separating the $\alpha$ and the other angle $90$ . In this case, by comparing you can see that $A \rightarrow \alpha$ and $B \rightarrow 90°$ . Substitute these quantities into the identity and then simplify as much as possible.

$\begin{aligned} \cos (α - 90 °) & = \cos α \cos 90 ° + \sin α \sin 90 ° \\ = k (0) + \sin α (1) \\ = 0 (k) + 1 \sin α \end{aligned}$ $\begin{aligned} \cos(\alpha -90°) & = \cos \alpha \cos 90° + \sin \alpha \sin 90° \\ & = k \left( 0 \right) + \sin\alpha \left( 1 \right) \\ & = \text{0}(k) + \text{1} \sin\alpha \end{aligned}$

The question states that you must write 'a' in place of $\alpha$ , so the final answer is: $\text{0}k + \text{1} \sin(a)$ .

Submit your answer as:

Trigonometric functions and their properties

Study the following graph and answer the following three questions:

What is the maximum value for this graph?

Answer: The maximum value is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is highest point on the graph before it starts sloping down?

STEP: Read the maximum value from the graph
[−1 point ⇒ 0 / 1 points left]

The maximum value of a graph is the highest $y$ -value reached by a graph. The $y$ -values cannot increase beyong this value. In this case the maximum value is: $\text{2}$ .

Submit your answer as:
What is the name of the trigonometric function represented by this graph? Assume that the graph has not been shifted in the horizontal direction.

INSTRUCTION: Select your answer from the drop down menu below.

Answer:
This curve represents a graph.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

This is either a sine or cosine graph. What makes a sine graph different from a cosine graph?

STEP: Identify the name of the function from the graph
[−1 point ⇒ 0 / 1 points left]

The graph in this question is a sine graph.

It is very easy to make a mistake that this is a cosine graph. We need to find the differences between a sine graph and a cosine graph.

The cosine graph has the same shape as the sine graph, but the cosine graph is shifted to the left by $90°$ .

A way to distinguish the two graphs is to look at the value of the $y$ -intercept. For the cosine graph, the $y$ -intercept is at the maximum $y$ -value. For the sine graph, the $y$ -intercept is halfway between the minimum and maximum values of $y$ . For this graph, the minimum $y$ -value is $-\text{4}$ . The maximum $y$ -value is $\text{2}$ . The $y$ -value of the coordinates of the $y$ -intercept is $-\text{1}$ . The $y$ -intercept is halfway between the maximum and minimum $y$ -values, so this is a sine graph.

Therefore the correct option is: sine.

Submit your answer as:
Give any value that lies within the range of the graph. (Your answer should be a number.)

Answer: A value that lies within the range is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at values between the maximum and minimum $y$ -values on the graph.

STEP: Identify the range of the function and choose any value within that range
[−1 point ⇒ 0 / 1 points left]

The range of a function is the set of all output values ( $y$ -values) we can get from a function when we are given a set of input values ( $x$ -values). In this case we know the maximum value is $\text{2}$ and the minimum value is $-\text{4}$ . Every $y$ -value from $-\text{4}$ to $\text{2}$ is in the range. So we can choose any value between $-\text{4}$ and $\text{2}$ .

The acceptable answer is any output value from $-\text{4}$ and $\text{2}$ .

Submit your answer as:

Properties of cosine functions

The diagram below shows the graph of $g(\theta) = a \cos \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the peak of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(\theta) = a \cos \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depths of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a peak: $(\text{0}\text{°}; \text{1})$ . We will substitute these coordinates into the equation of $g(\theta)$ and solve for $a$ .

$\begin{aligned} g (θ) & = a \cos θ \\ (1) & = a \cos (0 °) \\ 1 & = a (1) \\ 1 & = a \end{aligned}$ $\begin{aligned} g(\theta) &= a \cos \theta \\ (1) &= a \cos (0°) \\ 1 &= a (1) \\ 1 &= a \end{aligned}$

NOTE: If we look at the graph of $g(\theta)$ again, we see that the height of the peaks or depth of the troughs is $\text{1}$ . This means that the amplitude of $g(\theta)$ is $\text{1}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any cosine function of the form $f = a \cos \theta$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(\theta)$ : $g(\theta) = \cos \theta$ .

The value of $a$ is $\text{1}$ .

Submit your answer as:
The graph of $g$ is translated $\text{3}$ units upwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = \cos \theta + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = \cos \theta + q$ , where $q$ is the vertical shift. The value of $q$ is positive if we shift the graph upwards and negative if we shift the graph downwards.

The graph of $g$ has been shifted upwards by $\text{3}$ units. This means the value of $q$ is $\text{3}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ upwards by $\text{3}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $\theta$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 1$
- $q = 3$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (1) + (3) & \leq h \leq & (1) + (3) \\ 2 & \leq h \leq & 4 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(1) + (3) & \leq \; h \; \leq & (1) + (3) \\ 2 & \leq \; h \; \leq & 4 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $\text{2}$ and the maximum is $\text{4}$ .

The correct answer is: $2 \leq h \leq 4$ .

Submit your answer as:

Calculating the period of cosine graphs

Shown the following graph of the following form: $f(x) = \cos(kx)$ .

Calculate the value of $k$ (positive value).

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \cos (k x)

$f(x) = \cos(kx)$ with the period of

y = \cos (x)

$y = \cos(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \cos(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{180} \\ scale & = 2 \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{180}} \\ \textrm{scale} & = 2 \end{aligned}$

Therefore the horizontal scale factor is $2$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the $\cos(x)$ graph is symmetrical about the y-axis the sign of the variable $k$ has no effect on the shape of the graph. Therefore, the value of $k$ can be either $+ 2$ or $- 2$ .

NOTE: The final equation of the graph can be written as either:

f (x) = \cos (2 \times x)

$f(x) = \cos(2 \times x)$ or

f (x) = \cos (- 2 \times x)

$f(x) = \cos(-2 \times x)$

Because this question asks for the positive value, the correct answer is: $2$ .

Submit your answer as:

Exercises

Minimum and maximum values

Given a function

\begin{aligned} f (x) = 2 \cos x - 1 \end{aligned}

$\begin{aligned} f(x) = 2 \cos x - 1 \end{aligned}$

over the domain $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ .

What are the amplitude and range of $f(x)$ ?
Answer:
The amplitude of $f(x)$ is .

The range of $f(x)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \cos x + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(x)$ .

STEP: Determine the amplitude of $f(x)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \cos x + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \cos x + q$ is not affected by the value of $q$ .

For $f(x) = 2 \cos x - 1$ , the value of $a$ is $\text{2}$ . As a result, the amplitude of $f(x)$ is $\text{2}$ .

The amplitude of $f(x)$ is $\text{2}$ .

STEP: Determine the range of $f(x)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \cos x + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(x)$ .

In $f(x) = 2 \cos x - 1$ , the values and $A$ and $q$ are:
- $A = 2$
- $q = -1$
The range of $f(x)$ is:

$\begin{array}{rcl} - A + q & \leq f (x) \leq & A + q \\ - (2) + (- 1) & \leq f (x) \leq & (2) + (- 1) \\ - 3 & \leq f (x) \leq & 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(x) \leq & A + q \\ -(2) + (-1) & \leq \; f(x)\; \leq & (2) + (-1) \\ -3 & \leq \; f(x) \; \leq & 1 \end{array}$

The correct answers are:
- The amplitude is $\text{2}$ .
- The range is: $-3 \leq f(x) \leq 1$ .
Submit your answer as: and
What is the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ ?

Answer:
The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(x)$ and read off its maximum value within the given domain.

STEP: Sketch the graph of $f(x)$
[−2 points ⇒ 1 / 3 points left]

We need to find the maximum output value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ . To find this value, we need to sketch the graph of $f(x)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the maximum value from the graph.

The above figure is a sketch of the graph of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ . The maximum value has been marked with a dot.

STEP: Read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ directly from the graph. The coordinates of the maximum value are $(\text{0}\text{°}; \text{1})$ . The maximum value is the $y$ -value of the coordinates $(\text{0}\text{°}; \text{1})$ , that is, $\text{1}$ .

The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{210}\text{°}$ is $\text{1}$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (x) = - 3 \cos x + 2 \end{aligned}

$\begin{aligned} f(x) = -3 \cos x + 2 \end{aligned}$

over the domain $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ .

Determine the amplitude and range of $f(x)$ .
Answer:
The amplitude of $f(x)$ is .

The range of $f(x)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \cos x + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(x)$ .

STEP: Determine the amplitude of $f(x)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \cos x + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \cos x + q$ is not affected by the value of $q$ .

For $f(x) = -3 \cos x + 2$ , the value of $a$ is $-\text{3}$ . As a result, the amplitude of $f(x)$ is $\text{3}$ . We ignore the sign of $-\text{3}$ as we have mentioned.

The amplitude of $f(x)$ is $\text{3}$ .

STEP: Determine the range of $f(x)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \cos x + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(x)$ .

In $f(x) = -3 \cos x + 2$ , the values and $A$ and $q$ are:
- $A = 3$
- $q = 2$
The range of $f(x)$ is:

$\begin{array}{rcl} - A + q & \leq f (x) \leq & A + q \\ - (3) + (2) & \leq f (x) \leq & (3) + (2) \\ - 1 & \leq f (x) \leq & 5 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(x) \leq & A + q \\ -(3) + (2) & \leq \; f(x)\; \leq & (3) + (2) \\ -1 & \leq \; f(x) \; \leq & 5 \end{array}$

The correct answers are:
- The amplitude is $\text{3}$ .
- The range is: $-1 \leq f(x) \leq 5$ .
Submit your answer as: and
What is the minimum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ ?

Answer:
The minimum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(x)$ and read off its minimum value within the given domain.

STEP: Sketch the graph of $f(x)$
[−2 points ⇒ 1 / 3 points left]

We need to find the minimum output value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ . To find this value, we need to sketch the graph of $f(x)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the minimum value from the graph.

The above figure is a sketch of the graph of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ . The minimum value has been marked with a dot.

STEP: Read off the minimum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the minimum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ directly from the graph. The coordinates of the minimum value are $(\text{0}\text{°}; -\text{1})$ . The minimum value is the $y$ -value of the coordinates $(\text{0}\text{°}; -\text{1})$ , that is, $-\text{1}$ .

The minimum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{195}\text{°}$ is $-\text{1}$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (x) = 2 \cos x + 1 \end{aligned}

$\begin{aligned} f(x) = 2 \cos x + 1 \end{aligned}$

over the domain $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ .

What are the amplitude and range of $f(x)$ ?
Answer:
The amplitude of $f(x)$ is .

The range of $f(x)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \cos x + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(x)$ .

STEP: Determine the amplitude of $f(x)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \cos x + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \cos x + q$ is not affected by the value of $q$ .

For $f(x) = 2 \cos x + 1$ , the value of $a$ is $\text{2}$ . As a result, the amplitude of $f(x)$ is $\text{2}$ .

The amplitude of $f(x)$ is $\text{2}$ .

STEP: Determine the range of $f(x)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \cos x + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(x)$ .

In $f(x) = 2 \cos x + 1$ , the values and $A$ and $q$ are:
- $A = 2$
- $q = 1$
The range of $f(x)$ is:

$\begin{array}{rcl} - A + q & \leq f (x) \leq & A + q \\ - (2) + (1) & \leq f (x) \leq & (2) + (1) \\ - 1 & \leq f (x) \leq & 3 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(x) \leq & A + q \\ -(2) + (1) & \leq \; f(x)\; \leq & (2) + (1) \\ -1 & \leq \; f(x) \; \leq & 3 \end{array}$

The correct answers are:
- The amplitude is $\text{2}$ .
- The range is: $-1 \leq f(x) \leq 3$ .
Submit your answer as: and
What is the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ ?

Answer:
The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(x)$ and read off its maximum value within the given domain.

STEP: Sketch the graph of $f(x)$
[−2 points ⇒ 1 / 3 points left]

We need to find the maximum output value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ . To find this value, we need to sketch the graph of $f(x)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the maximum value from the graph.

The above figure is a sketch of the graph of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ . The maximum value has been marked with a dot.

STEP: Read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ directly from the graph. The coordinates of the maximum value are $(\text{0}\text{°}; \text{3})$ . The maximum value is the $y$ -value of the coordinates $(\text{0}\text{°}; \text{3})$ , that is, $\text{3}$ .

The maximum value of $f(x)$ for $\text{0}\text{°} \leq x \leq \text{225}\text{°}$ is $\text{3}$ .

Submit your answer as:

Finding a cosine equation from a graph.

The graph below shows a trigonometric equation of the following form: $y = a \cos x + q$ graphed over the domain $x \in \left[ 0° ; 360° \right]$ . Two points are shown on the graph: Point A at $\left(180°;\text{2} \right)$ , and point B at $\left(270°;\text{3} \right)$ .

Calculate the values of $a$ (the amplitude of the graph) and $q$ (the vertical shift of the graph).

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

a

$a$ .

STEP: Determine the amplitude of the curve
[−2 points ⇒ 2 / 4 points left]

There are some important points on a cosine graph. These are the maximum points, the minimum points, and the points in the middle. With those points we can figure out the values of $a$ and $q$ .

In this question, we have one point at the bottom and another point at the middle.

Let's start by finding the amplitude of the graph, which is the value of $a$ in the equation. The $y$ -value at the bottom is $\text{2}$ , while the $y$ -value at the the middle is $\text{3}$ . We can find the amplitude by working out the distance from the bottom of the graph to the middle: $\text{3} - (\text{2}) = 1$ .

At this point, we know that the equation is $y = \cos x + q$ .

STEP: Determine the vertical shift of the curve
[−2 points ⇒ 0 / 4 points left]

Now on to finding the value of $q$ . This can actually be done very quickly. Normally the sinusiodal axis sits at $y = 0$ . But the $q$ term moves that axis up or down. In this graph, the sinusoidal axis is at $y = 3$ ; the graph has been shifted up 3 spaces. Therefore $q = 3$ .

The equation for the graph shown is $y = \cos x+ 3$ .

The correct answers are $a = 1$ and $q = \text{3}$ .

Submit your answer as: and

Finding a cosine equation from a graph.

The graph below shows a trigonometric equation of the following form: $y = a \cos x + q$ graphed over the domain $x \in \left[ 0° ; 360° \right]$ . Two points are shown on the graph: Point A at $\left(90°;-\text{1} \right)$ , and point B at $\left(180°;-\text{2} \right)$ .

Calculate the values of $a$ (the amplitude of the graph) and $q$ (the vertical shift of the graph).

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

a

$a$ .

STEP: Determine the amplitude of the curve
[−2 points ⇒ 2 / 4 points left]

There are some important points on a cosine graph. These are the maximum points, the minimum points, and the points in the middle. With those points we can figure out the values of $a$ and $q$ .

In this question, we have one point at the middle and another point at the bottom.

Let's start by finding the amplitude of the graph, which is the value of $a$ in the equation. The $y$ -value at the middle is $-\text{1}$ , while the $y$ -value at the the bottom is $-\text{2}$ . We can find the amplitude by working out the distance from the middle of the graph to the bottom: $-\text{1} - (-\text{2}) = 1$ .

At this point, we know that the equation is $y = \cos x + q$ .

STEP: Determine the vertical shift of the curve
[−2 points ⇒ 0 / 4 points left]

Now on to finding the value of $q$ . This can actually be done very quickly. Normally the sinusiodal axis sits at $y = 0$ . But the $q$ term moves that axis up or down. In this graph, the sinusoidal axis is at $y = -1$ ; the graph has been shifted down 1 space. Therefore $q = -1$ .

The equation for the graph shown is $y = \cos x -1$ .

The correct answers are $a = 1$ and $q = \text{−1}$ .

Submit your answer as: and

Finding a cosine equation from a graph.

The graph below shows a trigonometric equation of the following form: $y = a \cos x + q$ graphed over the domain $x \in \left[ 0° ; 360° \right]$ . Two points are shown on the graph: Point A at $\left(180°;-\text{2,5} \right)$ , and point B at $\left(360°;\text{0,5} \right)$ .

Calculate the values of $a$ (the amplitude of the graph) and $q$ (the vertical shift of the graph).

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

a

$a$ .

STEP: Determine the amplitude of the curve
[−2 points ⇒ 2 / 4 points left]

There are some important points on a cosine graph. These are the maximum points, the minimum points, and the points in the middle. With those points we can figure out the values of $a$ and $q$ .

In this question, we have one point at the bottom and another point at the top.

Let's start by finding the amplitude of the graph, which is the value of $a$ in the equation. The $y$ -value at the bottom is $-\text{2,5}$ , while the $y$ -value at the the top is $\text{0,5}$ . The distance from the top to the bottom is $\text{0,5} - ( -\text{2,5} ) = 3$ . The amplitude is half of that. So the amplitude of this graph, and the value of $a$ , is $\frac{3}{2}$ .

At this point, we know that the equation is $y = \frac{3}{2} \cos x + q$ .

STEP: Determine the vertical shift of the curve
[−2 points ⇒ 0 / 4 points left]

Now on to finding the value of $q$ . This can actually be done very quickly. Normally the sinusiodal axis sits at $y = 0$ . But the $q$ term moves that axis up or down. In this graph, the sinusoidal axis is at $y = -1$ ; the graph has been shifted down 1 space. Therefore $q = -1$ .

The equation for the graph shown is $y = \frac{3}{2} \cos x -1$ .

The correct answers are $a = \frac{3}{2}$ and $q = \text{−1}$ .

Submit your answer as: and

Calculating the period of sine graphs

Shown the following graph of the following form: $f(x) = \sin(kx)$ .

Calculate the value of $k$ .

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \sin (k x)

$f(x) = \sin(kx)$ with the period of

y = \sin (x)

$y = \sin(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \sin(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{540} \\ scale & = \frac{2}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{540}} \\ \textrm{scale} & = \frac{2}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{2}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled and flipped across the y-axes, it must be negative!

The value of $k$ is $-\frac{2}{3}$ .

NOTE:The final equation of the graph is:

f (x) = \sin (- \frac{2}{3} \cdot x)

$f(x) = \sin(-\frac{2}{3} \cdot x)$

Submit your answer as:

Calculating the period of sine graphs

Shown the following graph of the following form: $f(x) = \sin(kx)$ .

Calculate the value of $k$ .

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \sin (k x)

$f(x) = \sin(kx)$ with the period of

y = \sin (x)

$y = \sin(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \sin(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{180} \\ scale & = 2 \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{180}} \\ \textrm{scale} & = 2 \end{aligned}$

Therefore the horizontal scale factor is $2$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $2$ .

NOTE:The final equation of the graph is:

f (x) = \sin (2 \cdot x)

$f(x) = \sin(2 \cdot x)$

Submit your answer as:

Calculating the period of sine graphs

Shown the following graph of the following form: $f(x) = \sin(kx)$ .

Calculate the value of $k$ .

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \sin (k x)

$f(x) = \sin(kx)$ with the period of

y = \sin (x)

$y = \sin(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \sin(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{540} \\ scale & = \frac{2}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{540}} \\ \textrm{scale} & = \frac{2}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{2}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $\frac{2}{3}$ .

NOTE:The final equation of the graph is:

f (x) = \sin (\frac{2}{3} \cdot x)

$f(x) = \sin(\frac{2}{3} \cdot x)$

Submit your answer as:

Finding a tangent equation from a graph

The graph below shows a tangent curve with an equation of the form $y = a \tan x + q$ . Two points are labelled on the curve: Point A is at $\left( 360° ; -1 \right)$ , and point B is at $\left( 225° ; - \frac{1}{2} \right)$ .

Find the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The tangent graph normally crosses through the origin,

(0 °; 0)

$(0°;0)$ . However, the value of

q

$q$ shifts the graph up or down. Use point

A

$A$ to figure out how much the graph has moved up or down. That tells you the value of

q

$q$ .

STEP: Read off the value of

q

$q$ from point

A

$A$
[−1 point ⇒ 2 / 3 points left]

The graph of tangent normally has $x$ -intercepts at $x = 0°, 180°$ and $360°$ . The $q$ term in the equation has the effect of shifting the graph up or down. We can use the position of point $A$ to figure out how much the graph moved up or down: since point $A$ is at $\left( 360° ; -1 \right)$ , the graph moved down by $1$ space. So $c = -1$ .

At this point we know that the equation is $y = a \tan x -1$ .

STEP: Substitute in the coordinates of a point and solve for

a

$a$
[−2 points ⇒ 0 / 3 points left]

\begin{aligned} y & = a \tan x - 1 \\ _{coordinates of point B}^{substitute in the} ⟶ (- \frac{1}{2}) & = a \tan (225 °) - 1 \\ - \frac{1}{2} & = a (1) - 1 \\ - \frac{1}{2} + 1 & = a \\ \frac{1}{2} & = a \end{aligned}

$\begin{aligned} y & = a \tan x -1 \\ ^{\text{substitute in the}}_{\text{coordinates of point }B} \quad \longrightarrow \quad \left( - \frac{1}{2} \right) & = a \tan \left( \text{225}° \right) -1 \\ - \frac{1}{2} & = a \left( \text{1} \right) -1 \\ - \frac{1}{2} + 1 & = a \\ \frac{1}{2} & = a \end{aligned}$

There we have it: the value of $a$ is $\frac{1}{2}$ . Putting it all together, we have the complete equation, which is $y = \frac{1}{2} \tan x -1$ .

The answers are $a = \frac{1}{2}$ and $q = -1$ .

Submit your answer as: and

Finding a tangent equation from a graph

Find the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The tangent graph normally crosses through the origin,

(0 °; 0)

$(0°;0)$ . However, the value of

q

$q$ shifts the graph up or down. Use point

A

$A$ to figure out how much the graph has moved up or down. That tells you the value of

q

$q$ .

STEP: Read off the value of

q

$q$ from point

A

$A$
[−1 point ⇒ 2 / 3 points left]

The graph of tangent normally has $x$ -intercepts at $x = 0°, 180°$ and $360°$ . The $q$ term in the equation has the effect of shifting the graph up or down. We can use the position of point $A$ to figure out how much the graph moved up or down: since point $A$ is at $\left( 360° ; -1 \right)$ , the graph moved down by $1$ space. So $c = -1$ .

At this point we know that the equation is $y = a \tan x -1$ .

STEP: Substitute in the coordinates of a point and solve for

a

$a$
[−2 points ⇒ 0 / 3 points left]

\begin{aligned} y & = a \tan x - 1 \\ _{coordinates of point B}^{substitute in the} ⟶ (1) & = a \tan (225 °) - 1 \\ 1 & = a (1) - 1 \\ 1 + 1 & = a \\ 2 & = a \end{aligned}

$\begin{aligned} y & = a \tan x -1 \\ ^{\text{substitute in the}}_{\text{coordinates of point }B} \quad \longrightarrow \quad \left( 1 \right) & = a \tan \left( \text{225}° \right) -1 \\ 1 & = a \left( \text{1} \right) -1 \\ 1 + 1 & = a \\ 2 & = a \end{aligned}$

There we have it: the value of $a$ is $2$ . Putting it all together, we have the complete equation, which is $y = 2 \tan x -1$ .

The answers are $a = 2$ and $q = -1$ .

Submit your answer as: and

Finding a tangent equation from a graph

On the graph below you see a tangent curve of the following form: $y = a \tan x + q$ . Two points are labelled on the curve: Point A is at $\left( 180° ; \frac{3}{2} \right)$ , and point B is at $\left( 45° ; \frac{7}{2} \right)$ .

Find $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The tangent graph normally crosses through the origin,

(0 °; 0)

$(0°;0)$ . However, the value of

q

$q$ shifts the graph up or down. Use point

A

$A$ to figure out how much the graph has moved up or down. That tells you the value of

q

$q$ .

STEP: Read off the value of

q

$q$ from point

A

$A$
[−1 point ⇒ 2 / 3 points left]

The graph of tangent normally has $x$ -intercepts at $x = 0°, 180°$ and $360°$ . The $q$ term in the equation has the effect of shifting the graph up or down. We can use the position of point $A$ to figure out how much the graph moved up or down: since point $A$ is at $\left( 180° ; \frac{3}{2} \right)$ , the graph moved up by $\frac{3}{2}$ spaces. So $c = \frac{3}{2}$ .

At this point we know that the equation is $y = a \tan x + \frac{3}{2}$ .

STEP: Substitute in the coordinates of a point and solve for

a

$a$
[−2 points ⇒ 0 / 3 points left]

\begin{aligned} y & = a \tan x + \frac{3}{2} \\ _{coordinates of point B}^{substitute in the} ⟶ (\frac{7}{2}) & = a \tan (45 °) + \frac{3}{2} \\ \frac{7}{2} & = a (1) + \frac{3}{2} \\ \frac{7}{2} - \frac{3}{2} & = a \\ 2 & = a \end{aligned}

$\begin{aligned} y & = a \tan x + \frac{3}{2} \\ ^{\text{substitute in the}}_{\text{coordinates of point }B} \quad \longrightarrow \quad \left( \frac{7}{2} \right) & = a \tan \left( \text{45}° \right) + \frac{3}{2} \\ \frac{7}{2} & = a \left( \text{1} \right) + \frac{3}{2} \\ \frac{7}{2} - \frac{3}{2} & = a \\ 2 & = a \end{aligned}$

There we have it: the value of $a$ is $2$ . Putting it all together, we have the complete equation, which is $y = 2 \tan x + \frac{3}{2}$ .

The answers are $a = 2$ and $q = \frac{3}{2}$ .

Submit your answer as: and

Finding a sine equation from a graph

The graph here shows an equation of the following form: $y = a \sin x + q$ . Point $A$ is at $(360°;\text{0,5})$ , and point $B$ is at $(90°;\text{2})$ . Find the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The value of $a$ is linked to the height of the sine curve, while the value of $q$ shifts the curve up or down.

STEP: Use the amplitude to find the value of

a

$a$
[−1 point ⇒ 1 / 2 points left]

In this question, we have one point at the middle and another point at the top of the sine curve. We can use those points to find both $a$ and $q$ .

Although we can find either of the two answers from the information given in the question, we will figure out the answer for $a$ first: $a$ is directly related to the amplitude of the curve. The $y$ -value at the middle is $\text{0,5}$ , while the $y$ -value at the top is $\text{2}$ . We can find the amplitude by working out the distance from the middle of the graph to the top: $\text{2} - (\text{0,5}) = \frac{3}{2}$ .

At this point, we have worked out half of the answer: the equation is $y = \frac{3}{2} \sin x + q$ .

STEP: Find

q

$q$ from the vertical shift of the curve
[−1 point ⇒ 0 / 2 points left]

Now we will figure out the value of $q$ . Remember that the sinusoidal axis for sine and cosine usually sits at $y = 0$ ; but the $q$ term moves that axis up or down. Look at the graph to see how the curve moved: in this graph, the sinusoidal axis is at $y = \frac{1}{2}$ ; the graph has been shifted up $\frac{1}{2}$ of a space. So $q = \frac{1}{2}$ .

The complete equation for the graph shown in this question is $y = \frac{3}{2} \sin x+ \frac{1}{2}$ .

The correct answers are:

$a = \frac{3}{2}$

$q = \frac{1}{2}$

Submit your answer as: and

Finding a sine equation from a graph

The graph here shows an equation of the following form: $y = a \sin x + q$ . Point $A$ is at $(90°;-\text{1,5})$ , and point $B$ is at $(360°;-\text{2})$ . Determine the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The value of $a$ is linked to the height of the sine curve, while the value of $q$ shifts the curve up or down.

STEP: Use the amplitude to find the value of

a

$a$
[−1 point ⇒ 1 / 2 points left]

In this question, we have one point at the top and another point at the middle of the sine curve. We can use those points to find both $a$ and $q$ .

Although we can find either of the two answers from the information given in the question, we will figure out the answer for $a$ first: $a$ is directly related to the amplitude of the curve. The $y$ -value at the top is $-\text{1,5}$ , while the $y$ -value at the middle is $\text{−2}$ . We can find the amplitude by working out the distance from the top of the graph to the middle: $-\text{1,5} - (-\text{2}) = \frac{1}{2}$ .

At this point, we have worked out half of the answer: the equation is $y = \frac{1}{2} \sin x + q$ .

STEP: Find

q

$q$ from the vertical shift of the curve
[−1 point ⇒ 0 / 2 points left]

Now we will figure out the value of $q$ . Remember that the sinusoidal axis for sine and cosine usually sits at $y = 0$ ; but the $q$ term moves that axis up or down. Look at the graph to see how the curve moved: in this graph, the sinusoidal axis is at $y = -2$ ; the graph has been shifted down $2$ spaces. So $q = -2$ .

The complete equation for the graph shown in this question is $y = \frac{1}{2} \sin x -2$ .

The correct answers are:

$a = \frac{1}{2}$

$q = -2$

Submit your answer as: and

Finding a sine equation from a graph

The graph here shows an equation of the following form: $y = a \sin x + q$ . Point $A$ is at $(270°;-\text{3})$ , and point $B$ is at $(0°;-\text{2})$ . Determine the values of $a$ and $q$ .

Answer:

$a =$

$q =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The value of $a$ is linked to the height of the sine curve, while the value of $q$ shifts the curve up or down.

STEP: Use the amplitude to find the value of

a

$a$
[−1 point ⇒ 1 / 2 points left]

In this question, we have one point at the bottom and another point at the middle of the sine curve. We can use those points to find both $a$ and $q$ .

Although we can find either of the two answers from the information given in the question, we will figure out the answer for $a$ first: $a$ is directly related to the amplitude of the curve. The $y$ -value at the bottom is $-\text{3}$ , while the $y$ -value at the middle is $\text{−2}$ . We can find the amplitude by working out the distance from the bottom of the graph to the middle: $-\text{2} - (-\text{3}) = 1$ .

At this point, we have worked out half of the answer: the equation is $y = \sin x + q$ .

STEP: Find

q

$q$ from the vertical shift of the curve
[−1 point ⇒ 0 / 2 points left]

Now we will figure out the value of $q$ . Remember that the sinusoidal axis for sine and cosine usually sits at $y = 0$ ; but the $q$ term moves that axis up or down. Look at the graph to see how the curve moved: in this graph, the sinusoidal axis is at $y = -2$ ; the graph has been shifted down $2$ spaces. So $q = -2$ .

The complete equation for the graph shown in this question is $y = \sin x -2$ .

The correct answers are:

$a = 1$

$q = -2$

Submit your answer as: and

Working with the period of tangent graphs

The graph below shows a tangent function of the following form: $f(x) = \tan(kx)$ . The asymptotes lie at $-\text{45}°$ and $\text{45}°$ .

Calculate the value of $k$ .

INSTRUCTION: Your answer must be either an integer or a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \tan (k x)

$f(x) = \tan(kx)$ with the period of

y = \tan (x)

$y = \tan(x)$
[−1 point ⇒ 2 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \tan(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{90}{45} \\ scale & = 2 \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{90}{\text{45}} \\ \textrm{scale} & = 2 \end{aligned}$

Therefore the horizontal scale factor is $2$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $2$ .

NOTE: The final equation of the graph is therefore:

f (x) = \tan (2 \cdot x)

$f(x) = \tan(2 \cdot x)$

Submit your answer as:

Working with the period of tangent graphs

The graph below shows a tangent function of the following form: $f(x) = \tan(kx)$ . The asymptotes lie at $-\text{45}°$ and $\text{45}°$ .

Calculate the value of $k$ .

INSTRUCTION: Your answer must be either an integer or a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \tan (k x)

$f(x) = \tan(kx)$ with the period of

y = \tan (x)

$y = \tan(x)$
[−1 point ⇒ 2 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \tan(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{90}{45} \\ scale & = 2 \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{90}{\text{45}} \\ \textrm{scale} & = 2 \end{aligned}$

Therefore the horizontal scale factor is $2$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled but NOT flipped across the y-axes, it must be positive!

The value of $k$ is $2$ .

NOTE: The final equation of the graph is therefore:

f (x) = \tan (2 \cdot x)

$f(x) = \tan(2 \cdot x)$

Submit your answer as:

Working with the period of tangent graphs

The graph below shows a tangent function of the following form: $f(x) = \tan(kx)$ . The asymptotes lie at $-\text{135}°$ and $\text{135}°$ .

Calculate the value of $k$ .

INSTRUCTION: Your answer must be either an integer or a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \tan (k x)

$f(x) = \tan(kx)$ with the period of

y = \tan (x)

$y = \tan(x)$
[−1 point ⇒ 2 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \tan(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{90}{135} \\ scale & = \frac{2}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{90}{\text{135}} \\ \textrm{scale} & = \frac{2}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{2}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the graph has been scaled and flipped across the y-axes, it must be negative!

The value of $k$ is $-\frac{2}{3}$ .

NOTE: The final equation of the graph is therefore:

f (x) = \tan (- \frac{2}{3} \cdot x)

$f(x) = \tan(-\frac{2}{3} \cdot x)$

Submit your answer as:

The cosine function

The table below shows input and output values for the trigonometric function $f(x) = \cos x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$?$
$45°$	$\frac{\sqrt{2}}{2}$
$90°$	$0$
$135°$	$-\frac{\sqrt{2}}{2}$
$180°$	$-1$
$225°$	$-\frac{\sqrt{2}}{2}$
$270°$	$0$
$315°$	$\frac{\sqrt{2}}{2}$
$360°$	$1$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \cos x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(0°)$ . So we need to evaluate the function with $x = 0°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(0°)$ is $1$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $1$ .

Submit your answer as:

The cosine function

The table below shows input and output values for the trigonometric function $f(x) = \cos x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$1$
$45°$	$?$
$90°$	$0$
$135°$	$-\frac{\sqrt{2}}{2}$
$180°$	$-1$
$225°$	$-\frac{\sqrt{2}}{2}$
$270°$	$0$
$315°$	$\frac{\sqrt{2}}{2}$
$360°$	$1$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

expression

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \cos x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(45°)$ . So we need to evaluate the function with $x = 45°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(45°)$ is $\frac{\sqrt{2}}{2}$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $\dfrac{\sqrt{2}}{2}$ .

Submit your answer as:

The cosine function

The table below shows input and output values for the trigonometric function $f(x) = \cos x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$1$
$60°$	$\frac{1}{2}$
$120°$	$-\frac{1}{2}$
$180°$	$-1$
$240°$	$-\frac{1}{2}$
$300°$	$?$
$360°$	$1$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \cos x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(300°)$ . So we need to evaluate the function with $x = 300°$ .

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $\dfrac{1}{2}$ .

Submit your answer as:

Properties of sine functions

The diagram below shows the graph of $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the peak of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depth of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a peak: $(\text{90}\text{°}; \text{1})$ . We will substitute these coordinates into the equation of $g(x)$ and solve for $a$ .

$\begin{aligned} g (x) & = a \sin x \\ (1) & = a \sin (90 °) \\ 1 & = a (1) \\ 1 & = a \end{aligned}$ $\begin{aligned} g(x) &= a \sin x \\ (1) &= a \sin (90°) \\ 1 &= a (1) \\ 1 &= a \end{aligned}$

NOTE: If we look at the graph of $g(x)$ again, we see that the height of the peaks or depth of the troughs is $\text{1}$ . This means that the amplitude of $g(x)$ is $\text{1}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any sine function of the form $f = a \sin x$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(x)$ : $g(x) = \sin x$ .

The value of $a$ is $\text{1}$ .

Submit your answer as:
The graph of $g$ is translated $\text{2}$ units downwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = \sin x + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = \sin x + q$ , where $q$ is the vertical shift. The value of $q$ is positive if the graph is shifted upwards and negative if the graph is shifted downwards.

The graph of $g$ has been shifted downwards by $\text{2}$ units. This means the value of $q$ is $-\text{2}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ downwards by $\text{2}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 1$
- $q = -2$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (1) + (- 2) & \leq h \leq & (1) + (- 2) \\ - 3 & \leq h \leq & - 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(1) + (-2) & \leq \; h \; \leq & (1) + (-2) \\ -3 & \leq \; h \; \leq & -1 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $-\text{3}$ and the maximum is $-\text{1}$ .

The correct answer is: $-3 \leq h \leq -1$ .

Submit your answer as:

Properties of sine functions

The diagram below shows the graph of $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the trough of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(x) = a \sin x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depth of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a trough: $(\text{90}\text{°}; -\text{2})$ . We will substitute these coordinates into the equation of $g(x)$ and solve for $a$ .

$\begin{aligned} g (x) & = a \sin x \\ (- 2) & = a \sin (90 °) \\ - 2 & = a (1) \\ - 2 & = a \end{aligned}$ $\begin{aligned} g(x) &= a \sin x \\ (-2) &= a \sin (90°) \\ -2 &= a (1) \\ -2 &= a \end{aligned}$

NOTE: If we look at the graph of $g(x)$ again, we see that the height of the peaks or depth of the troughs is $\text{2}$ . This means that the amplitude of $g(x)$ is $\text{2}$ . This is the same as the magnitude of $a$ that we have calculated. As a general rule, the amplitude of any sine function of the form $f = a \sin x$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(x)$ : $g(x) = -2 \sin x$ .

The value of $a$ is $-\text{2}$ .

Submit your answer as:
The graph of $g$ is translated $\text{3}$ units upwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = -2\sin x + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = -2\sin x + q$ , where $q$ is the vertical shift. The value of $q$ is positive if the graph is shifted upwards and negative if the graph is shifted downwards.

The graph of $g$ has been shifted upwards by $\text{3}$ units. This means the value of $q$ is $\text{3}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ upwards by $\text{3}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 2$
- $q = 3$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (2) + (3) & \leq h \leq & (2) + (3) \\ 1 & \leq h \leq & 5 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(2) + (3) & \leq \; h \; \leq & (2) + (3) \\ 1 & \leq \; h \; \leq & 5 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $\text{1}$ and the maximum is $\text{5}$ .

The correct answer is: $1 \leq h \leq 5$ .

Submit your answer as:

Properties of sine functions

The diagram below shows the graph of $g(\theta) = a \sin \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the peak of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(\theta) = a \sin \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depth of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a peak: $(\text{90}\text{°}; \text{4})$ . We will substitute these coordinates into the equation of $g(\theta)$ and solve for $a$ .

$\begin{aligned} g (θ) & = a \sin θ \\ (4) & = a \sin (90 °) \\ 4 & = a (1) \\ 4 & = a \end{aligned}$ $\begin{aligned} g(\theta) &= a \sin \theta \\ (4) &= a \sin (90°) \\ 4 &= a (1) \\ 4 &= a \end{aligned}$

NOTE: If we look at the graph of $g(\theta)$ again, we see that the height of the peaks or depth of the troughs is $\text{4}$ . This means that the amplitude of $g(\theta)$ is $\text{4}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any sine function of the form $f = a \sin \theta$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(\theta)$ : $g(\theta) = 4 \sin \theta$ .

The value of $a$ is $\text{4}$ .

Submit your answer as:
The graph of $g$ is translated $\text{2}$ units downwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 4\sin \theta + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 4\sin \theta + q$ , where $q$ is the vertical shift. The value of $q$ is positive if the graph is shifted upwards and negative if the graph is shifted downwards.

The graph of $g$ has been shifted downwards by $\text{2}$ units. This means the value of $q$ is $-\text{2}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ downwards by $\text{2}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $\theta$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 4$
- $q = -2$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (4) + (- 2) & \leq h \leq & (4) + (- 2) \\ - 6 & \leq h \leq & 2 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(4) + (-2) & \leq \; h \; \leq & (4) + (-2) \\ -6 & \leq \; h \; \leq & 2 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $-\text{6}$ and the maximum is $\text{2}$ .

The correct answer is: $-6 \leq h \leq 2$ .

Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function $f(x) = \tan x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$90°$	$\text{undefined}$
$180°$	$0$
$270°$	$?$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.

Answer: The missing value is .

string

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \tan x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(270°)$ . So we need to evaluate the function with $x = 270°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you can remember the answer) or using a calculator. In this case, if you use a calculator, you will get an error message. That is because $\tan(270°)$ does not have an answer. It is undefined. You can see this on the graph, because there is an asymptote at $x = 270°$ .

The asymptote comes from the fact that the tangent function is defined as $\frac{ y }{ x }$ on the Cartesian plane. If the angle is $90°$ or $270°$ , $x$ becomes zero. So the denominator becomes zero. And that is why the answer is undefined. (You can read more about this here.)

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is undefined.

Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function $f(x) = \tan x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$45°$	$1$
$90°$	$\text{undefined}$
$135°$	$-1$
$180°$	$0$
$225°$	$1$
$270°$	$\text{undefined}$
$315°$	$?$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \tan x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(315°)$ . So we need to evaluate the function with $x = 315°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(315°)$ is $-1$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $-1$ .

Submit your answer as:

The tangent function

The table below shows input and output values for the trigonometric function $f(x) = \tan x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$60°$	$\sqrt{3}$
$120°$	$-\sqrt{3}$
$180°$	$?$
$240°$	$\sqrt{3}$
$300°$	$-\sqrt{3}$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2. Type the word undefined if appropriate.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \tan x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(180°)$ . So we need to evaluate the function with $x = 180°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(180°)$ is $0$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $0$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (θ) = - 3 \sin θ - 2 \end{aligned}

$\begin{aligned} f(\theta) = -3 \sin \theta - 2 \end{aligned}$

over the domain $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ .

Determine the amplitude and range of $f(\theta)$ .
Answer:
The amplitude of $f(\theta)$ is .

The range of $f(\theta)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \sin \theta + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(\theta)$ .

STEP: Determine the amplitude of $f(\theta)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \sin \theta + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \sin \theta + q$ is not affected by the value of $q$ .

For $f(\theta) = -3 \sin \theta - 2$ , the value of $a$ is $-\text{3}$ . As a result, the amplitude of $f(\theta)$ is $\text{3}$ . We ignore the sign of $-\text{3}$ as we have mentioned.

The amplitude of $f(\theta)$ is $\text{3}$ .

STEP: Determine the range of $f(\theta)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \sin \theta + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(\theta)$ .

In $f(\theta) = -3 \sin \theta - 2$ , the values and $A$ and $q$ are:
- $A = 3$
- $q = -2$
The range of $f(\theta)$ is:

$\begin{array}{rcl} - A + q & \leq f (θ) \leq & A + q \\ - (3) + (- 2) & \leq f (θ) \leq & (3) + (- 2) \\ - 5 & \leq f (θ) \leq & 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(\theta) \leq & A + q \\ -(3) + (-2) & \leq \; f(\theta)\; \leq & (3) + (-2) \\ -5 & \leq \; f(\theta) \; \leq & 1 \end{array}$

The correct answers are:
- The amplitude is $\text{3}$ .
- The range is: $-5 \leq f(\theta) \leq 1$ .
Submit your answer as: and
What is the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ ?

Answer:
The minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(\theta)$ and read off its minimum value within the given domain.

STEP: Sketch the graph of $f(\theta)$
[−2 points ⇒ 1 / 3 points left]

We need to find the minimum output value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ . To find this value, we need to sketch the graph of $f(\theta)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the minimum value from the graph.

The above figure is a sketch of the graph of $f(\theta)$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ . The minimum value has been marked with a dot.

STEP: Read off the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ directly from the graph. The coordinates of the minimum value are $(\text{90}\text{°}; -\text{5})$ . The minimum value is the $y$ -value of the coordinates $(\text{90}\text{°}; -\text{5})$ , that is, $-\text{5}$ .

The minimum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{300}\text{°}$ is $-\text{5}$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (θ) = 3 \sin θ - 2 \end{aligned}

$\begin{aligned} f(\theta) = 3 \sin \theta - 2 \end{aligned}$

over the domain $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ .

Determine the amplitude and range of $f(\theta)$ .
Answer:
The amplitude of $f(\theta)$ is .

The range of $f(\theta)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \sin \theta + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(\theta)$ .

STEP: Determine the amplitude of $f(\theta)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \sin \theta + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \sin \theta + q$ is not affected by the value of $q$ .

For $f(\theta) = 3 \sin \theta - 2$ , the value of $a$ is $\text{3}$ . As a result, the amplitude of $f(\theta)$ is $\text{3}$ .

The amplitude of $f(\theta)$ is $\text{3}$ .

STEP: Determine the range of $f(\theta)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \sin \theta + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(\theta)$ .

In $f(\theta) = 3 \sin \theta - 2$ , the values and $A$ and $q$ are:
- $A = 3$
- $q = -2$
The range of $f(\theta)$ is:

$\begin{array}{rcl} - A + q & \leq f (θ) \leq & A + q \\ - (3) + (- 2) & \leq f (θ) \leq & (3) + (- 2) \\ - 5 & \leq f (θ) \leq & 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(\theta) \leq & A + q \\ -(3) + (-2) & \leq \; f(\theta)\; \leq & (3) + (-2) \\ -5 & \leq \; f(\theta) \; \leq & 1 \end{array}$

The correct answers are:
- The amplitude is $\text{3}$ .
- The range is: $-5 \leq f(\theta) \leq 1$ .
Submit your answer as: and
What is the maximum value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ ?

Answer:
The maximum value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(\theta)$ and read off its maximum value within the given domain.

STEP: Sketch the graph of $f(\theta)$
[−2 points ⇒ 1 / 3 points left]

We need to find the maximum output value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ . To find this value, we need to sketch the graph of $f(\theta)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the maximum value from the graph.

The above figure is a sketch of the graph of $f(\theta)$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ . The maximum value has been marked with a dot.

STEP: Read off the maximum value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the maximum value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ directly from the graph. The coordinates of the maximum value are $(\text{90}\text{°}; \text{1})$ . The maximum value is the $y$ -value of the coordinates $(\text{90}\text{°}; \text{1})$ , that is, $\text{1}$ .

The maximum value of $f(\theta)$ for $\text{60}\text{°} \leq \theta \leq \text{285}\text{°}$ is $\text{1}$ .

Submit your answer as:

Minimum and maximum values

Given a function

\begin{aligned} f (θ) = - 3 \sin θ - 1 \end{aligned}

$\begin{aligned} f(\theta) = -3 \sin \theta - 1 \end{aligned}$

over the domain $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ .

What are the amplitude and range of $f(\theta)$ ?
Answer:
The amplitude of $f(\theta)$ is .

The range of $f(\theta)$ is .

numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For a function $h = a \sin \theta + q$ , the amplitude of $h$ is the magnitude of $a$ without regard to its sign. For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Use this information to determine the range of $f(\theta)$ .

STEP: Determine the amplitude of $f(\theta)$
[−1 point ⇒ 1 / 2 points left]

The amplitude of any function $h = a \sin \theta + q$ , is the magnitude of $a$ without regard to its sign. This means the amplitude is always a positive number.

NOTE: The amplitude of any function of the form $h = a \sin \theta + q$ is not affected by the value of $q$ .

For $f(\theta) = -3 \sin \theta - 1$ , the value of $a$ is $-\text{3}$ . As a result, the amplitude of $f(\theta)$ is $\text{3}$ . We ignore the sign of $-\text{3}$ as we have mentioned.

The amplitude of $f(\theta)$ is $\text{3}$ .

STEP: Determine the range of $f(\theta)$
[−1 point ⇒ 0 / 2 points left]

For a function $h = a \sin \theta + q$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we can determine the range of $f(\theta)$ .

In $f(\theta) = -3 \sin \theta - 1$ , the values and $A$ and $q$ are:
- $A = 3$
- $q = -1$
The range of $f(\theta)$ is:

$\begin{array}{rcl} - A + q & \leq f (θ) \leq & A + q \\ - (3) + (- 1) & \leq f (θ) \leq & (3) + (- 1) \\ - 4 & \leq f (θ) \leq & 2 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; f(\theta) \leq & A + q \\ -(3) + (-1) & \leq \; f(\theta)\; \leq & (3) + (-1) \\ -4 & \leq \; f(\theta) \; \leq & 2 \end{array}$

The correct answers are:
- The amplitude is $\text{3}$ .
- The range is: $-4 \leq f(\theta) \leq 2$ .
Submit your answer as: and
What is the maximum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ ?

Answer:
The maximum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ is .

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Sketch the graph of $f(\theta)$ and read off its maximum value within the given domain.

STEP: Sketch the graph of $f(\theta)$
[−2 points ⇒ 1 / 3 points left]

We need to find the maximum output value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ . To find this value, we need to sketch the graph of $f(\theta)$ . Sketching graphs of functions is an important skill that is useful in mathematics. Once we have done that, we can read off the maximum value from the graph.

The above figure is a sketch of the graph of $f(\theta)$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The solid line is the part of the graph for the domain $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ . The maximum value has been marked with a dot.

STEP: Read off the maximum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$
[−1 point ⇒ 0 / 3 points left]

We can read off the maximum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ directly from the graph. The coordinates of the maximum value are $(\text{270}\text{°}; \text{2})$ . The maximum value is the $y$ -value of the coordinates $(\text{270}\text{°}; \text{2})$ , that is, $\text{2}$ .

The maximum value of $f(\theta)$ for $\text{75}\text{°} \leq \theta \leq \text{315}\text{°}$ is $\text{2}$ .

Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function $f(x) = \sin x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$30°$	$\frac{1}{2}$
$60°$	$\frac{\sqrt{3}}{2}$
$90°$	$1$
$120°$	$\frac{\sqrt{3}}{2}$
$150°$	$?$
$180°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \sin x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(150°)$ . So we need to evaluate the function with $x = 150°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(150°)$ is $\frac{1}{2}$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $\dfrac{1}{2}$ .

Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function $f(x) = \sin x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$60°$	$\frac{\sqrt{3}}{2}$
$120°$	$\frac{\sqrt{3}}{2}$
$180°$	$0$
$240°$	$-\frac{\sqrt{3}}{2}$
$300°$	$-\frac{\sqrt{3}}{2}$
$360°$	$?$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \sin x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(360°)$ . So we need to evaluate the function with $x = 360°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(360°)$ is $0$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $0$ .

Submit your answer as:

The sine function

The table below shows input and output values for the trigonometric function $f(x) = \sin x$ . One of the output values is missing.

$x$	$f(x)$
$0°$	$0$
$45°$	$\frac{\sqrt{2}}{2}$
$90°$	$1$
$135°$	$\frac{\sqrt{2}}{2}$
$180°$	$0$
$225°$	$-\frac{\sqrt{2}}{2}$
$270°$	$-1$
$315°$	$?$
$360°$	$0$

Determine the missing output value.

INSTRUCTION: Your answer must be exact - do not round off. If the answer includes a root, type it using sqrt. For example: sqrt(3)/2.

Answer: The missing value is .

expression

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use your calculator to find the answer.

STEP: Write down the answer or use your calculator
[−1 point ⇒ 0 / 1 points left]

We have the function $f(x) = \sin x$ and we need to fill in the missing value from the table. The missing value corresponds to $f(315°)$ . So we need to evaluate the function with $x = 315°$ .

The trigonometric functions are special functions which we can either evaluate from memory (if you remember the answer) or using a calculator. The value of $f(315°)$ is $-\frac{\sqrt{2}}{2}$ . This corresponds to the point shown in the graph below.

TIP: It is important to know how to use your calculator to calculate trigonometric values. Make sure that you are comfortable using your calculator to evaluate the trigonometric functions.

Therefore the missing value is $-\dfrac{\sqrt{2}}{2}$ .

Submit your answer as:

Trigonometry: interpreting graphs

The graphs show the functions $f(x) = c\cos x + d$ and $g(x) = a\tan x + b$ . Points $A( \text{45}; \text{2,0})$ and $B( \text{0}; \text{4})$ are labelled, and Point $N (x;y)$ is on $g(x)$ .

Find the numerical values of $a$ and $b$ for $g(x)$ .
Answer:
1. $a =$
2. $b =$
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Begin by identifying which graph is $f(x)$ and which is $g(x)$ . What are the effects of $a$ and $b$ on the graph?

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

It is often easiest to first find the vertical shift of the graph, $b$ . Examine the graph of $g(x)$ and determine if it has gone up or down. The graph in question is shown below highlighted in blue, and the centre line is shown as well. The centre line has not moved up or down from the $x$ -axis. Therefore, the value of $b$ is $\text{0}$ .

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The value of $a$ affects the amplitude: either stretching or flattening the graph. You can find the amplitude by looking at the peaks and valleys of the graph. For $\tan x$ it is important to remember that the normal graph passes through the point $(45° ; 1)$ . If the $\tan$ graph is multiplied by a coefficient other than one, it will move that point up or down by stretching or squeezing it. Making these comparisons, you should find that $a = 2.$

Submit your answer as: and
What is the minimum value of $g(x) -1$ ?

Answer: The answer is .
expression

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Questions about 'minimum' or 'maximum' values are about the lowest and highest points on a graph (the smallest and biggest $y$ -values). Think about the effect of the $-1$ on the graph: will it move the graph horizontally or vertically?

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

This question asks you to think about the graph of $g(x)$ when it has been changed due to the extra term of $-1$ . This term moves the entire graph down $-1$ space(s). In this case, you are moving the $\tan$ function, which goes down forever and up forever (along the asymptotes): the minimum value is $-\infty$ and the maximum value is $\infty$ . If the entire graph moves down $-1$ space(s), the graph still reaches down to $-\infty$ and up to $\infty$ . Therefore, the minimum value of $g(x) -1$ is: $-\infty$ .

Submit your answer as:
Find the coordinates of Point $N$ if it is $\text{+30}$ degrees from $A$ .

INSTRUCTION: Write your answer as a coordinate pair, with the values inside brackets and separated by a semi-colon ;. Round non-integer values to two decimal places.

Answer:
coordinate

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Point $N$ lies on the graph of $g(x)$ , so the coordinates must satisfy the equation for $g(x)$ . Use the function to work out the $y$ -coordinate. However, you will first need to determine the $x$ -coordinate.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

Point $N$ is $\text{+30}$ degrees from Point $A$ , which means the $x$ -coordinate must be $45° +30° = 75°$ . To find the $y$ -coordinate, substitute into the function. You already found the values of $a$ and $b$ ; so the function becomes: $g(x) = 2 \tan(x) + 0$ . Now substitute $x = 75°$ at Point $N$ and use your calculator to find the final answer: $\Rightarrow 2 \tan(75°) + 0 = 7.46$ . Therefore, the coordinates at Point $N$ are $( 75; 7.46)$ .

Submit your answer as:
Given $\cos(\alpha) = k$ , find a simplified expression for $\cos(\alpha +90°)$ in terms of $\alpha$ and $k$ .

INSTRUCTION: Type a in the place of $\alpha$ , and round your answer to two decimal places if there are non-integer values.

Answer: $\cos(\alpha +90°) =$
expression

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
This question is about compound angles: you need to use the compound angle identity which matches the question to expand the expression and get an answer.

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Simplifying the expression in the question requires using the compound angle identity $\cos \left( A + B\right) = \cos A \cos B - \sin A \sin B$ to expand the function by separating the $\alpha$ and the other angle $90$ . In this case, by comparing you can see that $A \rightarrow \alpha$ and $B \rightarrow 90°$ . Substitute these quantities into the identity and then simplify as much as possible.

$\begin{aligned} \cos (α + 90 °) & = \cos α \cos 90 ° - \sin α \sin 90 ° \\ = k (0) - \sin α (1) \\ = 0 (k) - 1 \sin α \end{aligned}$ $\begin{aligned} \cos(\alpha +90°) & = \cos \alpha \cos 90° - \sin \alpha \sin 90° \\ & = k \left( 0 \right) - \sin\alpha \left( 1 \right) \\ & = \text{0}(k) - \text{1} \sin\alpha \end{aligned}$

The question states that you must write 'a' in place of $\alpha$ , so the final answer is: $\text{0}k - \text{1} \sin(a)$ .

Submit your answer as:

Trigonometry: interpreting graphs

The graphs show the functions $f(x) = a\sin x + b$ and $g(x) = c\cos x + d$ . Points $A( \text{0}; \text{5,0})$ and $B( \text{90}; \text{3})$ are labelled, and Point $N (x;y)$ is on $f(x)$ .

Find the numerical values of $a$ and $b$ for $f(x)$ .
Answer:
1. $a =$
2. $b =$
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Begin by identifying which graph is $f(x)$ and which is $g(x)$ . What are the effects of $a$ and $b$ on the graph?

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

It is often easiest to first find the vertical shift of the graph, $b$ . Examine the graph of $f(x)$ and determine if it has gone up or down. The graph in question is shown below highlighted in blue, and the centre line is shown as well. The centre line has moved up from the $x$ -axis by $\text{1}$ space(s). Therefore, the value of $b$ is $\text{1}$ .

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The value of $a$ affects the amplitude: either stretching or flattening the graph. You can find the amplitude by looking at the peaks and valleys of the graph. For $\tan x$ it is important to remember that the normal graph passes through the point $(45° ; 1)$ . If the $\tan$ graph is multiplied by a coefficient other than one, it will move that point up or down by stretching or squeezing it. Making these comparisons, you should find that $a = 2.$

Submit your answer as: and
What is the minimum value of $g(x) -3$ ?

Answer: The answer is .
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Questions about 'minimum' or 'maximum' values are about the lowest and highest points on a graph (the smallest and biggest $y$ -values). Think about the effect of the $-3$ on the graph: will it move the graph horizontally or vertically?

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

This question asks you to think about the graph of $g(x)$ when it has been changed due to the extra term of $-3$ . This term moves the entire graph down $-3$ space(s). If you look at the graph of $g(x)$ above and imagine it moving down , then the lowest points will move to $y = -6$ and the highest points will end up at $y = 2$ . Therefore, the minimum value of $g(x) -3$ is: $-6$ .

Submit your answer as:
Determine the range of $f(x) -3$ .

INSTRUCTION: Write your answer in interval notation.

Answer: $y \in$
interval

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The range of a graph is the total 'height' of the graph from the smallest $y$ -value up to the maximum $y$ -value.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The range of the graph is the interval from the lowest $y$ -value up to the largest $y$ -value. In the case of $f(x) -3$ , it is the graph of $f(x)$ shifted down by $\text{3}$ space(s). You can see on the graph that $f(x)$ has a minimum value of $y = -1$ and a maximum value of $y = 3$ . When the graph is shifted due to the $-3$ , the range is the interval $[ -4; 0 ]$ .

Submit your answer as:
Given $\sin(\alpha) = k$ , find a simplified expression for $\sin(\alpha -60°)$ in terms of $\alpha$ and $k$ .

INSTRUCTION: Type a in the place of $\alpha$ , and round your answer to two decimal places if there are non-integer values.

Answer: $\sin(\alpha -60°) =$
expression

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
This question is about compound angles: you need to use the compound angle identity which matches the question to expand the expression and get an answer.

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Simplifying the expression in the question requires using the compound angle identity $\sin \left(A - B \right) = \sin A \cos B - \cos A \sin B$ to expand the function by separating the $\alpha$ and the other angle $60$ . In this case, by comparing you can see that $A \rightarrow \alpha$ and $B \rightarrow 60°$ . Substitute these quantities into the identity and then simplify as much as possible.

$\begin{aligned} \sin (α - 60 °) & = \sin α \cos 60 ° - \cos α \sin 60 ° \\ = k (\frac{1}{2}) - \cos α (\frac{\sqrt{3}}{2}) \\ = 0,5 (k) - 0,87 \cos α \end{aligned}$ $\begin{aligned} \sin(\alpha -60°) & = \sin \alpha \cos 60° - \cos \alpha \sin 60° \\ & = k \left( \frac{1}{2} \right) - \cos\alpha \left( \frac{\sqrt{3}}{2} \right) \\ & = \text{0,5}(k) - \text{0,87} \cos\alpha \end{aligned}$

The question states that you must write 'a' in place of $\alpha$ , so the final answer is: $\text{0,5}k - \text{0,87} \cos(a)$ .

Submit your answer as:

Trigonometry: interpreting graphs

The graphs show the functions $f(x) = a\sin x + b$ and $g(x) = c\tan x + d$ . Points $A( \text{45}; \text{0,5})$ and $B( \text{90}; \text{3})$ are labelled, and Point $Q (x;y)$ is on $g(x)$ .

Find the numerical values of $a$ and $b$ for $f(x)$ .
Answer:
1. $a =$
2. $b =$
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
Begin by identifying which graph is $f(x)$ and which is $g(x)$ . What are the effects of $a$ and $b$ on the graph?

STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

It is often easiest to first find the vertical shift of the graph, $b$ . Examine the graph of $f(x)$ and determine if it has gone up or down. The graph in question is shown below highlighted in blue, and the centre line is shown as well. The centre line has moved up from the $x$ -axis by $\text{2}$ space(s). Therefore, the value of $b$ is $\text{2}$ .

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

The value of $a$ affects the amplitude: either stretching or flattening the graph. You can find the amplitude by looking at the peaks and valleys of the graph. For $\tan x$ it is important to remember that the normal graph passes through the point $(45° ; 1)$ . If the $\tan$ graph is multiplied by a coefficient other than one, it will move that point up or down by stretching or squeezing it. Making these comparisons, you should find that $a = 1.$

Submit your answer as: and
Determine the period of $f\left( \frac{1}{4}x \right)$ .

Answer: The answer is .
numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

This question refers to the period, which is about how wide the graph is. Start with the period of the normal function, $f(x)$ . Then you need to adjust the period because the coefficient of $\frac{1}{4}$ changes the period of the graph by stretching or squeezing it side-to-side.

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

This question asks you to think about the graph of $f(x)$ when it has been changed due to the factor of $\frac{1}{4}$ multiplying $x$ . Calculating the period of $f\left( \frac{1}{4}x \right)$ requires dividing the normal period of the function by $\frac{1}{4}$ . The function $f(x)$ typically has a period of $360°$ , so the new period for $f\left( \frac{1}{4}x \right)$ is: $\left(360°\right) \div \left( \frac{1}{4} \right) = 1440°$ .

Submit your answer as:
Write down the $x$ -values of the first two positive asymptotes to the right of the $y$ -axis for $g\left( \frac{1}{4}x \right)$ .

INSTRUCTION: Give the two answers (without degrees) separated by a ;.

Answer: $x =$ $°$
list

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The asymptotes are vertical lines so you must identify the positions of the asymptotes on the $x$ -axis for the graph of $g\left( \frac{1}{4}x \right)$

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The asymptotes for $\tan(x)$ are at $x = \pm 90°, x = \pm 270°$ , etc. When the value of $x$ is multiplied, as in $g\left( \frac{1}{4}x \right)$ , the graph is stretched or squeezed to the sides. Just like for the period, the new positions of the asymptotes can be found by dividing the normal values by the coefficient of $x$ . Therefore, the $x$ values for the first two asymptotes are: $90° \div \frac{1}{4} = 360°$ and, $270° \div \frac{1}{4} = 1080°$

Submit your answer as:
Given $\sin(\alpha) = k$ , find a simplified expression for $\sin(\alpha -30°)$ in terms of $\alpha$ and $k$ .

INSTRUCTION: Type a in the place of $\alpha$ , and round your answer to two decimal places if there are non-integer values.

Answer: $\sin(\alpha -30°) =$
expression

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
This question is about compound angles: you need to use the compound angle identity which matches the question to expand the expression and get an answer.

STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

Simplifying the expression in the question requires using the compound angle identity $\sin \left(A - B \right) = \sin A \cos B - \cos A \sin B$ to expand the function by separating the $\alpha$ and the other angle $30$ . In this case, by comparing you can see that $A \rightarrow \alpha$ and $B \rightarrow 30°$ . Substitute these quantities into the identity and then simplify as much as possible.

$\begin{aligned} \sin (α - 30 °) & = \sin α \cos 30 ° - \cos α \sin 30 ° \\ = k (\frac{\sqrt{3}}{2}) - \cos α (\frac{1}{2}) \\ = 0,87 (k) - 0,5 \cos α \end{aligned}$ $\begin{aligned} \sin(\alpha -30°) & = \sin \alpha \cos 30° - \cos \alpha \sin 30° \\ & = k \left( \frac{\sqrt{3}}{2} \right) - \cos\alpha \left( \frac{1}{2} \right) \\ & = \text{0,87}(k) - \text{0,5} \cos\alpha \end{aligned}$

The question states that you must write 'a' in place of $\alpha$ , so the final answer is: $\text{0,87}k - \text{0,5} \cos(a)$ .

Submit your answer as:

Trigonometric functions and their properties

Study the following graph and answer the following three questions:

What is the maximum value for this graph?

Answer: The maximum value is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is highest point on the graph before it starts sloping down?

STEP: Read the maximum value from the graph
[−1 point ⇒ 0 / 1 points left]

The maximum value of a graph is the highest $y$ -value reached by a graph. The $y$ -values cannot increase beyong this value. In this case the maximum value is: $\text{3}$ .

Submit your answer as:
What is the name of the trigonometric function represented by this graph? Assume that the graph has not been shifted in the horizontal direction.

INSTRUCTION: Select your answer from the drop down menu below.

Answer:
This curve represents a graph.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

This is either a sine or cosine graph. What makes a sine graph different from a cosine graph?

STEP: Identify the name of the function from the graph
[−1 point ⇒ 0 / 1 points left]

The graph in this question is a cosine graph.

It is very easy to make a mistake that this is a sine graph. We need to find the differences between a cosine graph and a sine graph.

The cosine graph has the same shape as the sine graph, but the cosine graph is shifted to the left by $90°$ .

A way to distinguish the two graphs is to look at the value of the $y$ -intercept. For the cosine graph, the $y$ -intercept is at the maximum $y$ -value. For the sine graph, the $y$ -intercept is halfway between the minimum and maximum values of $y$ . For this graph, the minimum $y$ -value is $-\text{3}$ . The maximum $y$ -value is $\text{3}$ . The $y$ -value of the coordinates of the $y$ -intercept is $\text{3}$ . The $y$ -intercept is at the maximum $y$ -value, so this is a cosine graph.

Therefore the correct option is: cosine.

Submit your answer as:
Give any value that lies within the range of the graph. (Your answer should be a number.)

Answer: A value that lies within the range is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at values between the maximum and minimum $y$ -values on the graph.

STEP: Identify the range of the function and choose any value within that range
[−1 point ⇒ 0 / 1 points left]

The range of a function is the set of all output values ( $y$ -values) we can get from a function when we are given a set of input values ( $x$ -values). In this case we know the maximum value is $\text{3}$ and the minimum value is $-\text{3}$ . Every $y$ -value from $-\text{3}$ to $\text{3}$ is in the range. So we can choose any value between $-\text{3}$ and $\text{3}$ .

The acceptable answer is any output value from $-\text{3}$ and $\text{3}$ .

Submit your answer as:

Trigonometric functions and their properties

Study the following graph and answer the following three questions:

What is the maximum value for this graph?

Answer: The maximum value is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is highest point on the graph before it starts sloping down?

STEP: Read the maximum value from the graph
[−1 point ⇒ 0 / 1 points left]

The maximum value of a graph is the highest $y$ -value reached by a graph. The $y$ -values cannot increase beyong this value. In this case the maximum value is: $\text{5}$ .

Submit your answer as:
What is the name of the trigonometric function represented by this graph? Assume that the graph has not been shifted in the horizontal direction.

INSTRUCTION: Select your answer from the drop down menu below.

Answer:
This curve represents a graph.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

This is either a sine or cosine graph. What makes a sine graph different from a cosine graph?

STEP: Identify the name of the function from the graph
[−1 point ⇒ 0 / 1 points left]

The graph in this question is a sine graph.

It is very easy to make a mistake that this is a cosine graph. We need to find the differences between a sine graph and a cosine graph.

The cosine graph has the same shape as the sine graph, but the cosine graph is shifted to the left by $90°$ .

A way to distinguish the two graphs is to look at the value of the $y$ -intercept. For the cosine graph, the $y$ -intercept is at the maximum $y$ -value. For the sine graph, the $y$ -intercept is halfway between the minimum and maximum values of $y$ . For this graph, the minimum $y$ -value is $-\text{1}$ . The maximum $y$ -value is $\text{5}$ . The $y$ -value of the coordinates of the $y$ -intercept is $\text{2}$ . The $y$ -intercept is halfway between the maximum and minimum $y$ -values, so this is a sine graph.

Therefore the correct option is: sine.

Submit your answer as:
Give any value that lies within the range of the graph. (Your answer should be a number.)

Answer: A value that lies within the range is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at values between the maximum and minimum $y$ -values on the graph.

STEP: Identify the range of the function and choose any value within that range
[−1 point ⇒ 0 / 1 points left]

The range of a function is the set of all output values ( $y$ -values) we can get from a function when we are given a set of input values ( $x$ -values). In this case we know the maximum value is $\text{5}$ and the minimum value is $-\text{1}$ . Every $y$ -value from $-\text{1}$ to $\text{5}$ is in the range. So we can choose any value between $-\text{1}$ and $\text{5}$ .

The acceptable answer is any output value from $-\text{1}$ and $\text{5}$ .

Submit your answer as:

Trigonometric functions and their properties

Study the following graph and answer the following three questions:

What is the maximum value for this graph?

Answer: The maximum value is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

What is highest point on the graph before it starts sloping down?

STEP: Read the maximum value from the graph
[−1 point ⇒ 0 / 1 points left]

The maximum value of a graph is the highest $y$ -value reached by a graph. The $y$ -values cannot increase beyong this value. In this case the maximum value is: $\text{1}$ .

Submit your answer as:
What is the name of the trigonometric function represented by this graph? Assume that the graph has not been shifted in the horizontal direction.

INSTRUCTION: Select your answer from the drop down menu below.

Answer:
This curve represents a graph.

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

This is either a sine or cosine graph. What makes a sine graph different from a cosine graph?

STEP: Identify the name of the function from the graph
[−1 point ⇒ 0 / 1 points left]

The graph in this question is a sine graph.

It is very easy to make a mistake that this is a cosine graph. We need to find the differences between a sine graph and a cosine graph.

The cosine graph has the same shape as the sine graph, but the cosine graph is shifted to the left by $90°$ .

A way to distinguish the two graphs is to look at the value of the $y$ -intercept. For the cosine graph, the $y$ -intercept is at the maximum $y$ -value. For the sine graph, the $y$ -intercept is halfway between the minimum and maximum values of $y$ . For this graph, the minimum $y$ -value is $-\text{5}$ . The maximum $y$ -value is $\text{1}$ . The $y$ -value of the coordinates of the $y$ -intercept is $-\text{2}$ . The $y$ -intercept is halfway between the maximum and minimum $y$ -values, so this is a sine graph.

Therefore the correct option is: sine.

Submit your answer as:
Give any value that lies within the range of the graph. (Your answer should be a number.)

Answer: A value that lies within the range is .
numeric

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at values between the maximum and minimum $y$ -values on the graph.

STEP: Identify the range of the function and choose any value within that range
[−1 point ⇒ 0 / 1 points left]

The range of a function is the set of all output values ( $y$ -values) we can get from a function when we are given a set of input values ( $x$ -values). In this case we know the maximum value is $\text{1}$ and the minimum value is $-\text{5}$ . Every $y$ -value from $-\text{5}$ to $\text{1}$ is in the range. So we can choose any value between $-\text{5}$ and $\text{1}$ .

The acceptable answer is any output value from $-\text{5}$ and $\text{1}$ .

Submit your answer as:

Properties of cosine functions

The diagram below shows the graph of $g(x) = a \cos x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the trough of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(x) = a \cos x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depths of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a trough: $(\text{180}\text{°}; -\text{2})$ . We will substitute these coordinates into the equation of $g(x)$ and solve for $a$ .

$\begin{aligned} g (x) & = a \cos x \\ (- 2) & = a \cos (180 °) \\ - 2 & = a (- 1) \\ 2 & = a \end{aligned}$ $\begin{aligned} g(x) &= a \cos x \\ (-2) &= a \cos (180°) \\ -2 &= a (-1) \\ 2 &= a \end{aligned}$

NOTE: If we look at the graph of $g(x)$ again, we see that the height of the peaks or depth of the troughs is $\text{2}$ . This means that the amplitude of $g(x)$ is $\text{2}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any cosine function of the form $f = a \cos x$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(x)$ : $g(x) = 2 \cos x$ .

The value of $a$ is $\text{2}$ .

Submit your answer as:
The graph of $g$ is translated $\text{2}$ units upwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = 2\cos x + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = 2\cos x + q$ , where $q$ is the vertical shift. The value of $q$ is positive if we shift the graph upwards and negative if we shift the graph downwards.

The graph of $g$ has been shifted upwards by $\text{2}$ units. This means the value of $q$ is $\text{2}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ upwards by $\text{2}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 2$
- $q = 2$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (2) + (2) & \leq h \leq & (2) + (2) \\ 0 & \leq h \leq & 4 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(2) + (2) & \leq \; h \; \leq & (2) + (2) \\ 0 & \leq \; h \; \leq & 4 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $\text{0}$ and the maximum is $\text{4}$ .

The correct answer is: $0 \leq h \leq 4$ .

Submit your answer as:

Properties of cosine functions

The diagram below shows the graph of $g(\theta) = a \cos \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the peak of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(\theta) = a \cos \theta$ for $\text{0}\text{°} \leq \theta \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depths of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a peak: $(\text{0}\text{°}; \text{3})$ . We will substitute these coordinates into the equation of $g(\theta)$ and solve for $a$ .

$\begin{aligned} g (θ) & = a \cos θ \\ (3) & = a \cos (0 °) \\ 3 & = a (1) \\ 3 & = a \end{aligned}$ $\begin{aligned} g(\theta) &= a \cos \theta \\ (3) &= a \cos (0°) \\ 3 &= a (1) \\ 3 &= a \end{aligned}$

NOTE: If we look at the graph of $g(\theta)$ again, we see that the height of the peaks or depth of the troughs is $\text{3}$ . This means that the amplitude of $g(\theta)$ is $\text{3}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any cosine function of the form $f = a \cos \theta$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(\theta)$ : $g(\theta) = 3 \cos \theta$ .

The value of $a$ is $\text{3}$ .

Submit your answer as:
The graph of $g$ is translated $\text{2}$ units downwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 3\cos \theta + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ downwards. This means the equation of $h$ has the form $h = 3\cos \theta + q$ , where $q$ is the vertical shift. The value of $q$ is positive if we shift the graph upwards and negative if we shift the graph downwards.

The graph of $g$ has been shifted downwards by $\text{2}$ units. This means the value of $q$ is $-\text{2}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ downwards by $\text{2}$ units.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $\theta$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 3$
- $q = -2$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (3) + (- 2) & \leq h \leq & (3) + (- 2) \\ - 5 & \leq h \leq & 1 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(3) + (-2) & \leq \; h \; \leq & (3) + (-2) \\ -5 & \leq \; h \; \leq & 1 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $-\text{5}$ and the maximum is $\text{1}$ .

The correct answer is: $-5 \leq h \leq 1$ .

Submit your answer as:

Properties of cosine functions

The diagram below shows the graph of $g(x) = a \cos x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ .

Determine the value of $a$ .

Answer:
The value of $a$ is .

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Read off the coordinates of the peak or the trough from the graph. Then substitute these coordinates into the equation and solve for $a$ .

STEP: Read off the coordinates of the trough of the graph and use them to solve for $a$
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $a$ in $g(x) = a \cos x$ for $\text{0}\text{°} \leq x \leq \text{360}\text{°}$ . The value of $a$ affects the amplitude of the graph, that is, the height of the peaks and the depths of the troughs. So we need to read off the coordinates of either a peak or a trough of the graph and use them to find the value of $a$ .

NOTE: We could use the coordinates of any point except those that are on the $x$ -axis. For this graph, it is easier to read off the coordinates of either the peak or the trough. Can you think of why we do not use the points on the $x$ -axis?

In this solution, we will use the coordinates of a trough: $(\text{180}\text{°}; -\text{4})$ . We will substitute these coordinates into the equation of $g(x)$ and solve for $a$ .

$\begin{aligned} g (x) & = a \cos x \\ (- 4) & = a \cos (180 °) \\ - 4 & = a (- 1) \\ 4 & = a \end{aligned}$ $\begin{aligned} g(x) &= a \cos x \\ (-4) &= a \cos (180°) \\ -4 &= a (-1) \\ 4 &= a \end{aligned}$

NOTE: If we look at the graph of $g(x)$ again, we see that the height of the peaks or depth of the troughs is $\text{4}$ . This means that the amplitude of $g(x)$ is $\text{4}$ . This is the same value that we have calculated for $a$ . As a general rule, the amplitude of any cosine function of the form $f = a \cos x$ , is the positive value of $a$ .

With this value of $a$ , we can now write the equation of $g(x)$ : $g(x) = 4 \cos x$ .

The value of $a$ is $\text{4}$ .

Submit your answer as:
The graph of $g$ is translated $\text{1}$ unit upwards to get a new graph $h$ . What is the range of $h$ ?
Answer:
The range of $h$ is .
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The graph of $h$ results from shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = 4\cos x + q$ , where $q$ is the vertical shift. Start by determining the value of $q$ . Then to determine the range, use the fact that $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ .

STEP: Determine the value of $q$
[−1 point ⇒ 1 / 2 points left]

The graph of $h$ results from translating or shifting the graph of $g$ upwards. This means the equation of $h$ has the form $h = 4\cos x + q$ , where $q$ is the vertical shift. The value of $q$ is positive if we shift the graph upwards and negative if we shift the graph downwards.

The graph of $g$ has been shifted upwards by $\text{1}$ unit. This means the value of $q$ is $\text{1}$ .

The figure below shows a plot of the graphs of the functions $g$ and $h$ on the same axes.

We can now see the effect of shifting the graph of $g$ upwards by $\text{1}$ unit.

STEP: Determine the range of $h$
[−1 point ⇒ 0 / 2 points left]

For all values of $x$ , $h$ is always between $-A + q$ and $A + q$ , where $A$ is the amplitude of $h$ . Using this information, we now determine the range of $h$ .

The values of $A$ and $q$ for the graph $h$ are:
- $A = 4$
- $q = 1$
The range of $h$ is:

$\begin{array}{rcl} - A + q & \leq h \leq & A + q \\ - (4) + (1) & \leq h \leq & (4) + (1) \\ - 3 & \leq h \leq & 5 \end{array}$ $\begin{array}{rcl} -A + q & \leq \; h \leq & A + q \\ -(4) + (1) & \leq \; h \; \leq & (4) + (1) \\ -3 & \leq \; h \; \leq & 5 \end{array}$

This matches the range we can see on the graph perfectly: the minimum value of $h$ is $-\text{3}$ and the maximum is $\text{5}$ .

The correct answer is: $-3 \leq h \leq 5$ .

Submit your answer as:

Calculating the period of cosine graphs

Shown the following graph of the following form: $f(x) = \cos(kx)$ .

Calculate the value of $k$ (positive value).

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \cos (k x)

$f(x) = \cos(kx)$ with the period of

y = \cos (x)

$y = \cos(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \cos(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{270} \\ scale & = \frac{4}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{270}} \\ \textrm{scale} & = \frac{4}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{4}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the $\cos(x)$ graph is symmetrical about the y-axis the sign of the variable $k$ has no effect on the shape of the graph. Therefore, the value of $k$ can be either $+ \frac{4}{3}$ or $- \frac{4}{3}$ .

NOTE: The final equation of the graph can be written as either:

f (x) = \cos (\frac{4}{3} \times x)

$f(x) = \cos(\frac{4}{3} \times x)$ or

f (x) = \cos (- \frac{4}{3} \times x)

$f(x) = \cos(-\frac{4}{3} \times x)$

Because this question asks for the positive value, the correct answer is: $\frac{4}{3}$ .

Submit your answer as:

Calculating the period of cosine graphs

Shown the following graph of the following form: $f(x) = \cos(kx)$ .

Calculate the value of $k$ (positive value).

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \cos (k x)

$f(x) = \cos(kx)$ with the period of

y = \cos (x)

$y = \cos(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \cos(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{540} \\ scale & = \frac{2}{3} \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{540}} \\ \textrm{scale} & = \frac{2}{3} \end{aligned}$

Therefore the horizontal scale factor is $\frac{2}{3}$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the $\cos(x)$ graph is symmetrical about the y-axis the sign of the variable $k$ has no effect on the shape of the graph. Therefore, the value of $k$ can be either $+ \frac{2}{3}$ or $- \frac{2}{3}$ .

NOTE: The final equation of the graph can be written as either:

f (x) = \cos (\frac{2}{3} \times x)

$f(x) = \cos(\frac{2}{3} \times x)$ or

f (x) = \cos (- \frac{2}{3} \times x)

$f(x) = \cos(-\frac{2}{3} \times x)$

Because this question asks for the positive value, the correct answer is: $\frac{2}{3}$ .

Submit your answer as:

Calculating the period of cosine graphs

Shown the following graph of the following form: $f(x) = \cos(kx)$ .

Calculate the value of $k$ (positive value).

INSTRUCTION: Write your answer as a fraction.

Answer:

k =

$k =$

fraction

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

First we recall the fact that variable $k$ affects the period of the waveform.

STEP: Compare the period of

f (x) = \cos (k x)

$f(x) = \cos(kx)$ with the period of

y = \cos (x)

$y = \cos(x)$
[−1 point ⇒ 2 / 3 points left]

To understand the effect that $k$ has on the waveform, we can compare the given graph to the standard form of the graph: $y = \cos(x)$

We can measure the relative change in the period by measuring the periods of each graph (shown above as $G$ and $H$ ).

STEP: Calculate the horizontal scale factor
[−1 point ⇒ 1 / 3 points left]

To calculate the value of $k$ , we want to calculate "how much" has the period of the waveform decreased? So let us calculate the scale factor:

\begin{aligned} H \times scale & = G \\ scale & = \frac{G}{H} \\ scale & = \frac{360}{180} \\ scale & = 2 \end{aligned}

$\begin{aligned} H \times \textrm{scale} & = G \\ \textrm{scale} & = \frac{G}{H} \\ \textrm{scale} & = \frac{360}{\text{180}} \\ \textrm{scale} & = 2 \end{aligned}$

Therefore the horizontal scale factor is $2$ .

STEP: Determine the value of

k

$k$
[−1 point ⇒ 0 / 3 points left]

Finally we ask: Is the value of $k$ positive or negative?

Because the $\cos(x)$ graph is symmetrical about the y-axis the sign of the variable $k$ has no effect on the shape of the graph. Therefore, the value of $k$ can be either $+ 2$ or $- 2$ .

NOTE: The final equation of the graph can be written as either:

f (x) = \cos (2 \times x)

$f(x) = \cos(2 \times x)$ or

f (x) = \cos (- 2 \times x)

$f(x) = \cos(-2 \times x)$

Because this question asks for the positive value, the correct answer is: $2$ .

Submit your answer as:

Trigonometry

1. Trigonometric graphs

Minimum and maximum values

Finding a cosine equation from a graph.

Calculating the period of sine graphs

Finding a tangent equation from a graph

Finding a sine equation from a graph

Working with the period of tangent graphs

The cosine function

Properties of sine functions

The tangent function

Minimum and maximum values

The sine function

Trigonometry: interpreting graphs

Trigonometric functions and their properties

Properties of cosine functions

Calculating the period of cosine graphs

Exercises

Minimum and maximum values

Minimum and maximum values

Minimum and maximum values

Finding a cosine equation from a graph.

Finding a cosine equation from a graph.

Finding a cosine equation from a graph.

Calculating the period of sine graphs

Calculating the period of sine graphs

Calculating the period of sine graphs

Finding a tangent equation from a graph

Finding a tangent equation from a graph

Finding a tangent equation from a graph

Finding a sine equation from a graph

Finding a sine equation from a graph

Finding a sine equation from a graph

Working with the period of tangent graphs

Working with the period of tangent graphs

Working with the period of tangent graphs

The cosine function

The cosine function

The cosine function

Properties of sine functions

Properties of sine functions

Properties of sine functions

The tangent function

The tangent function

The tangent function

Minimum and maximum values

Minimum and maximum values

Minimum and maximum values

The sine function

The sine function

The sine function

Trigonometry: interpreting graphs

Trigonometry: interpreting graphs

Trigonometry: interpreting graphs

Trigonometric functions and their properties

Trigonometric functions and their properties

Trigonometric functions and their properties

Properties of cosine functions

Properties of cosine functions

Properties of cosine functions

Calculating the period of cosine graphs

Calculating the period of cosine graphs

Calculating the period of cosine graphs

2. Trigonometric equations

Exercises